I'm thinking about results of the form: Under assumption $A$, if $X$ is a compact Hausdorff space and $C$ is a cover of $X$ by closed $G\_\delta$ sets, then there is a subcover of cardinality $\leq\kappa$.


Obviously in general we can take $\kappa\leq|X|$ and in some cases this is the best we can do such as in Cantor space in which every point is a closed $G\_\delta$ set.


The strongest result I've found so far is this: If $X$ is a scattered compact Hausdorff space and $C$ is a cover of $X$ by closed $G\_\delta$ sets, then there is a countable subcover.


Proof: Since $X$ is scattered it is totally disconnected, and we can represent each $F\in C$ as $F=\bigcap\_{n<\omega}Q\_{F,n}$, where $Q\_{F,n}$ is a nested sequence of clopen sets. Partition $X$ into a finite collection $X\_0, \dots ,X\_{n-1}$ of clopen sets such that each $X\_i$ has a unique point $x\_i$ such that $CB(x\_i) = CB(X)$ (where $CB$ is the Cantor-Bendixson rank). In each $X\_i$ choose some $F\_i\in C$ such that $x\_i \in F$ and then consider $X\_i \setminus F\_i$, which can be written as a countable union of clopen sets. Iterate this procedure in each such clopen set. This forms a well-founded forest which is countably branching and thus countable. The subcover is the $F\_i$'s chosen at each node of the forest.


This is certainly optimal for the general scattered case, as you can construct a cover that saturates the bound in any infinite scattered compact Hausdorff space.


A weak generalization is this: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G\_\delta$ sets, then for any $C\_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq \omega^{|C\_0| + 1}$.


Proof: If $P$ is empty we can use the previous result, so assume that $P$ is non-empty. Let $C\_0$ be a cover of $P$, so in particular $P \subseteq \bigcup\_{F\in C\_0}\bigcap\_{n<\omega}U\_{F,n} = \bigcap\_{g:C\_0\rightarrow\omega}\bigcup\_{F\in C\_0}U\_{F,g(F)}$. By compactness, for each $g:C\_0 \rightarrow \omega$, there is a finite set $C\_{0,g}\subseteq C\_0$ such that $P\subseteq \bigcup\_{F\in C\_{0,g}}U\_{F,g(F)}$. So let $R\_g=X\setminus \bigcup\_{F\in C\_{0,g}}U\_{F,g(F)}$. Each $R\_g$ is closed and scattered as a subspace of $X$, so we can run the previous argument to get a countable $C\_g \subseteq C$ that covers $R\_g$. This means that in order to cover the rest of $X$ we need at most $\omega^{|C\_0|}\times \omega = \omega^{|C\_0|+1}$ more sets, so since $\omega^{|C\_0|+1} \geq |C\_0|$ we need at most $\omega^{|C\_0|+1}$ all together.


This is only clearly non-trivial in cases where $P$ is much smaller than $X$. The question is how much better can you do? In particular is there a universal bound for all compact Hausdorff spaces? Can you get a better bound in terms of $|C\_0|$? Can you get bounds in terms of other cardinal invariants of $P$ or $X$, such as the weight or density character? Can you say anything non-trivial about perfect spaces? What is $\kappa$ for $\beta \omega \setminus \omega$ or $2^\lambda$ for various cardinals $\lambda$?


EDIT: In the negative direction I suspect you can show that if $X$ is a compact Hausdorff space with non-empty perfect core, then $X$ has a cover by closed $G\_\delta$ sets such that no subcover has cardinality $< 2^\omega$.


EDIT2: Suppose that $X$ is a compact Hausdorff space with non-empty perfect core $P$. Construct a tree of closed sets $\{F\_{\sigma}\}\_{\sigma \in 2^{<\omega}}$ with the following properties: Every $F^\circ\_{\sigma}$ has non-empty intersection with $P$. For every $\sigma \in 2^{<\omega}$, $F\_{\sigma\frown 0},F\_{\sigma\frown 1} \subset F\_\sigma^\circ$ and $F\_{\sigma\frown 0}\cap F\_{\sigma \frown 1} = \varnothing$. (We can do this since $P$ is perfect.) By these conditions, for any $\alpha \in 2^\omega$, the set $G\_\alpha = \bigcap\_{n<\omega} F\_{\alpha \upharpoonright n}$ is a closed $G\_\delta$ set. If $\alpha,\beta \in 2^\omega$ with $\alpha \neq \beta$, then $G\_\alpha \cap G\_\beta = \varnothing$, and $H = \bigcup\_{\alpha \in 2^\omega} G\_\alpha$ is a closed $G\_\delta$ set. Since $H$ is a closed $G\_\delta$ set, we can find a sequence of closed $G\_\delta$ sets $\{D\_n\}\_{n<\omega}$ such that $\bigcup\_{n<\omega} D\_n = X \setminus H$. So then the cover $\{D\_n\}\_{n<\omega} \cup \{ G\_\alpha \}\_{\alpha \in 2^\omega}$ is a cover by closed $G\_\delta$ sets with no subcover of cardinality $< 2^\omega$.


EDIT3: There's a much better bound in the second case that I missed: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G\_\delta$ sets, then for any $C\_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq |C\_0|+\omega$.


Proof: Same proof as before, just notice that while there are $\omega^{|C\_0|}$ many functions $g:C\_0 \rightarrow \omega$, there are only $(|C\_0|\times \omega)^{<\omega} = |C\_0|+\omega$ many finite subsets of $\{U\_{F,n}\}\_{F\in C,n<\omega}$, so there are only at most $|C\_0|+\omega$ many $R\_g$'s to consider, and we get that $C$ has a subcover of size at most $|C\_0|+(|C\_0|+\omega)\times\omega = |C\_0|+\omega$.


By the same argument as in the scattered case, this is basically optimal relative to the size of $|C\_0|$, so now the questions really is about perfect spaces.


EDIT4: Combining the content of Taras and Anonymous's comments, you get a more refined bound in terms of generalized Cantor-Bendixson ranks (these probably have a name somewhere), specifically for any cardinal $\kappa$ we can can define 


$$X^\prime\_{\kappa}=X \setminus \bigcup \{U\subseteq X : U\text{ open, has density character }\leq\kappa\}$$


$$X^{(\alpha)}\_{\kappa} = \bigcap\_{\beta < \alpha} (X^{(\beta)}\_{\kappa})^\prime \_{\kappa}$$


where $\bigcap\varnothing=X$, for any $\alpha \in \text{Ord}\cup\{\infty\}$. And $CB\_{\kappa}(X)$ is the smallest ordinal $\alpha$ for which $X^{(\alpha + 1)}\_{\kappa}=\varnothing$, if it exists, and $\infty$ otherwise. Then if we let $S(X)$ be the smallest $\kappa$ such that $CB\_{\kappa}(X)<\infty$, we should get a better bound in terms of $S(X)$.


If the density character of $X$ is $\kappa$, then every closed $G\_\delta$ cover has a subcover of cardinality $\leq 2^\kappa$, since there are only at most that many closed $G\_\delta$ sets in $X$.


Assume that for some cardinal $\kappa$ and some ordinal $\alpha$, we've shown that if $CB\_{\kappa}(X)<\alpha$, then every closed $G\_\delta$ cover has a subcover of cardinality $\leq 2^\lambda$ for some $\lambda < \kappa$. Let $X$ be a compact Hausdorff space such that $CB\_{\kappa}(X) = \alpha$ and let $C$ be a cover of $X$ by closed $G\_\delta$ sets. This implies that the density character of $P = X^{(CB\_{\kappa}(X))}\_{\kappa}$ is $\leq\kappa$, so there is some set $C\_0 \subseteq C$ of cardinality $\leq 2^\kappa$ such that $C\_0$ covers $P$. By the same argument as before, $X\setminus \bigcup C\_0$ can be written as the union of at most $2^\kappa$ many closed subspaces, each of which has strictly smaller $CB\_{\kappa}$. So by the induction hypothesis each of these has a subset of $C$ of cardinality $\leq 2^\kappa$ that covers it, so overall $X$ is covered by a subset of $C$ of cardinality $\leq 2^\kappa$.


So by induction, if $CB\_{\kappa}(X)<\infty$, then every closed $G\_\delta$ cover of $X$ has a subcover of cardinality $\leq 2^\kappa$. So in general any closed $G\_\delta$ cover of a compact Hausdorff space has a subcover of cardinality $\leq 2^{S(X)}$.


You can probably get slightly tighter bounds at singular cardinals using a CB derivative for density characters $<\kappa$ rather than just $\leq \kappa$.



Covering compact Hausdorff spaces with closed $G_\delta$ sets
Your question is related to a pair of old questions of A.V.Arhangel'skii. First of all note that in a regular space, for every point $x$ and every $G\_\delta$ set G containing $x$ there is a closed $G\_\delta$ $H$ contained in $G$ such that $x \in H$. So the topology generated by the closed $G\_\delta$ sets of a compact Hausdorff space $X$ coincides with the "$G\_\delta$-topology" $X\_\delta$ (that is, the topology generated by the $G\_\delta$ subsets of $X$). 


The *Lindelof degree* of $X$ ($L(X)$) is defined as the minimum cardinal $\kappa$ such that every open cover of $X$ has a $\leq \kappa$-sized subcover and the *weak Lindelof degree of $X$* ($wL(X)$) is defined as the minimum cardinal $\kappa$ such that every open cover of $X$ has a $\leq \kappa$ subcollection whose union is dense in $X$.



> 
> Question (A.V. Arhangel'skii, 1970): Let $X$ be a compact Hausdorff space.
> 
> 
> 1. Is it true that $L(X\_\delta) \leq 2^{\aleph\_0}$?
> 2. Is it true that $wL(X\_\delta) \leq 2^{\aleph\_0}$?
> 
> 
> 


Various partial positive answers to these questions can be found in the literature. For example (see the references below):


* In 1972 Juhász proved that $wL(X\_\delta) \leq 2^{\aleph\_0}$ for every compact ccc space $X$.
* In 1974 Fleischmann and Williams proved that $L(X\_\delta) \leq 2^{\aleph\_0}$ for every compact linearly ordered space $X$.
* In 1985 Pytkeev proved that $L(X\_\delta) \leq 2^{\aleph\_0}$ for every compact space of countable tightness $X$.
* In 2016 I proved that $wL(X\_\delta) \leq 2^{\aleph\_0}$ for every compact space $X$ where player II has a winning strategy in the "weak Lindelof game of length $\omega\_1$" (which among other things implies Juhász's 1972 result).


However both questions have a negative answer. A construction of a compact space $X$ such that $wL(X\_\delta) > 2^{\aleph\_0}$ can be found in my paper with Szeptycki (item 5 from the reference list). To find a compact space $X$ such that $L(X\_\delta) > 2^{\aleph\_0}$ it's enough to take the Cantor cube $X=2^{\mathfrak{c}^+}$ (just note that $X$ is homeomorphic to $(2^\omega)^{\mathfrak{c}^+}$, so $X\_\delta$ is homeomorphic to the $G\_\delta$ topology on $D^{\mathfrak{c}^+}$, where $D$ is a discrete space of size continuum and a $G\_\delta$ cover without subcovers of size continuum for the latter space is given by $\{[\sigma]: \sigma \in Fn(\mathfrak{c}^+, D, \omega\_1) \wedge \sigma$ is not one-to one on its domain $\}$ where $[\sigma]=\{f \in D^{\mathfrak{c}^+}: f \supset \sigma \}$).


Finally, to address your question regarding the existence of a bound, Toshimichi Usuba very recently proved that an $\omega\_1$-strongly compact cardinal is a *precise* upper bound on both the Lindelof degree and the weak Lindelof degree of the $G\_\delta$-topology on a compact space. So it's consistent that there is no bound to $wL(X\_\delta)$ for $X$ compact!


References:


1. *Juhász, István*, [**On two problems of A.V.Arkhangel’skij**](http://dx.doi.org/10.1016/0016-660X(72)90002-5), General Topology Appl. 2, 151-156 (1972). [ZBL0237.54002](https://zbmath.org/?q=an:0237.54002).
2. *Williams, Scott; Fleischman, William*, [**The $G\_\delta$-topology on compact spaces**](http://dx.doi.org/10.4064/fm-83-2-143-149), Fundam. Math. 83 (1974), pp. 143-149. [ZBL0278.54021](https://zbmath.org/?q=an:0278.54021).
3. *Pytkeev, E. G.*, About the $G\_\lambda$-topology and the power of some families of subsets on compacta, Topology theory and applications, 5th Colloq., Eger/Hung. 1983, Colloq. Math. Soc. János Bolyai 41, 517-522 (1985). [ZBL0604.54007](https://zbmath.org/?q=an:0604.54007).
4. *Spadaro, Santi*, [**Infinite games and chain conditions**](http://dx.doi.org/10.4064/fm232-3-2016), Fundam. Math. 234 (2016), pp. 229-239. [ZBL1360.54010](https://zbmath.org/?q=an:1360.54010).
5. *Spadaro, S.; Szeptycki, P.*, [**$G\_{\delta}$ covers of compact spaces**](http://dx.doi.org/10.1007/s10474-017-0785-4), Acta Math. Hung., 154 (2018) pp. 252-263. [ZBL06850215](https://zbmath.org/?q=an:06850215).
6. *Usuba, Toshimichi*, [**$G\_\delta $-topology and compact cardinals**](http://dx.doi.org/10.4064/fm487-7-2018), [ZBL07053981](https://zbmath.org/?q=an:07053981).
[1]: Fundam. Math. 246 (2019), 71--87.




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P.S.: A related question is: "how big can a partition of a compact space by closed $G\_\delta$s be?". In this case the continuum is a bound (see my answer to the following Mathoverflow question: [A generalization of the Arhangelskii Theorem](https://mathoverflow.net/questions/173523/a-generalization-of-the-arhangelskii-theorem/176221#176221)).


