Let $(X,\tau)$ be a topological space. We say ${\cal U}\subseteq \tau$ is an open cover if 


* $\bigcup {\cal U} = X$, and
* $X\notin {\cal U}$.


${\cal U}$ is *minimal* if for all $U\_0\in {\cal U}$ we have $\bigcup \big({\cal U}\setminus \{U\_0\}\big) \neq X$. Clearly, every $T\_1$-space on more than $1$ point possesses a minimal cover: pick $x\neq y\in X$ and let ${\cal U} = \big\{X\setminus\{x\}, X\setminus\{y\}\big\}$.


**Question.** Given any open cover of a Hausdorff space $(X,\tau)$ with $|X|>1$, does it have a [refinement](https://en.wikipedia.org/wiki/Cover_(topology)#Refinement) that is a minimal cover? 


**Note.** I did not ask for subcovers in the question because of the following example: Let $X = \mathbb{R}$ with the Euclidean topology, then ${\cal U} = \big\{ \{x\in\mathbb{R}: x < n\} : n\in\mathbb{N}, n\geq 1 \big\}$ does not have a minimal subcover.



Minimal refinements of open covers of $T_2$-spaces
No -- the space $\omega\_1$ (the first uncountable ordinal with the order topology) is a counterexample.


Consider the open cover $\mathcal V$ of $\omega\_1$ consisting of all bounded open sets. I claim that this cover has no minimal refinement. The proof is an application of [Fodor's pressing down lemma](https://en.wikipedia.org/wiki/Fodor%27s_lemma).


Suppose $\mathcal U$ is a refinement of $\mathcal V$, which just means that $\mathcal U$ is an open cover of $\omega\_1$ consisting of bounded sets. We'll show that $\mathcal U$ is not minimal. For each limit ordinal $\alpha$, choose some $U\_\alpha \in \mathcal U$ with $\alpha \in U\_\alpha$. Because $\alpha$ is a limit, there is some $\beta\_\alpha < \alpha$ such that 
$$(\beta\_\alpha,\alpha] \subseteq U\_\alpha.$$
The function $\alpha \mapsto \beta\_\alpha$ is regressive (i.e., it satisfies the hypothesis of Fodor's lemma), so there is some fixed $\beta$ and some uncountable set $S \subseteq \omega\_1$ such that $\beta\_\alpha = \beta$ for all $\alpha \in S$.


The interval $[0,\beta]$ is countable, so there is some $\alpha\_0 \in S$ such that
$$\bigcup \{[0,\beta] \cap U\_\alpha : \alpha \in S\} = \bigcup \{[0,\beta] \cap U\_\alpha : \alpha \in S, \alpha < \alpha\_0\}.$$
(In other words, any ordinal $\leq\!\beta$ that appears in some $U\_\alpha$, $\alpha \in S$, must appear by $\alpha\_0$.) Because $U\_{\alpha\_0}$ is bounded and $S$ is uncountable, there is some $\alpha\_1 \in S$ such that $\sup U\_{\alpha\_0} < \alpha\_1$.


The set $U\_{\alpha\_0}$ can be broken into two pieces: $U\_{\alpha\_0} \cap [0,\beta]$ and $U\_{\alpha\_0} \cap (\beta,\omega\_1)$. The first piece is contained in $\bigcup\{U\_\alpha : \alpha \in S, \alpha < \alpha\_0\}$, by our choice of $\alpha\_0$. The second piece is contained in $(\beta,\alpha\_1] \subseteq U\_{\alpha\_1}$, by our choice of $\alpha\_1$. Both pieces are covered by $\mathcal U \setminus \{U\_{\alpha\_0}\}$. Thus $\mathcal U \setminus \{U\_{\alpha\_0}\}$ is still an open cover of $\omega\_1$, and $\mathcal U$ is not minimal.


