Let $Y$ be a subset of a locally compact Hausdorff topological space $X$ and consider the following properties.


1. $\overline{Y}$ is compact.
2. Every open cover of $X$ has a finite subcover of $Y$.


Certainly 1. implies 2. Does 2. imply 1.?


If $X$ were also second countable, $X$ would metrizable and the answer would be yes.


If $X$ did not have to be locally compact, the answer would be no. To see this, extend the usual subspace topology on the closed unit ball $B$ of $\mathbb{R}^2$ by adding in sets of the form $\{x\}\cup(N\cap B^\circ)$, where $N$ is an open neighbourhood of $x$ in $\mathbb{R}^2$ and $x\in B\setminus B^\circ$. Then $B^\circ$ satisfies 2. but not 1.


Note: In my original question, 1. was instead "every net in $Y$ has a cluster point in $X$". As Nik pointed out, proving 2. implies 1. is then a simple exercise in topology. 



Relative Compactness vs Way Below in Locally Compact Hausdorff Spaces
Tristan pointed out in the comments that my argument showing (2) $\Rightarrow$ (1) in general is faulty. Still, I think it's true for locally compact spaces. Suppose $X$ is locally compact, $Y \subseteq X$, and every open cover of $X$ has a finite subcover of $Y$. Consider the covering of $X$ by all open subsets of whose closure is compact. Since finitely many of them cover $Y$, this implies that $\overline{Y}$ is compact.


My answer to the original question follows.


Let $(x\_\alpha)$ be a net in $Y$ and suppose every open cover of $X$ has a finite subcover of $Y$. For each $\alpha$ let $F\_\alpha$ be the closure, in $X$, of $\{x\_\beta: \beta \geq \alpha\}$, and let $U\_\alpha = X \setminus F\_\alpha$. If $\{U\_\alpha\}$ were an open cover of $X$ then by hypothesis there would be a finite subcover of $Y$, and then by directedness there would be a single $U\_\alpha$ containing $Y$, which is absurd. So $\{U\_\alpha\}$ cannot be an open cover of $X$, hence there exists $x \in X$ not in any $U\_\alpha$, i.e., $x \in F\_\alpha$ for all $\alpha$. So $x$ is a cluster point of the net.


