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

- $\overline{Y}$ is compact.
- 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.