Let $X$ be a locally compact Hausdorff space, and $A \subset X$ closed. I want to show that $X - A$ is locally compact.
I have found a proof here: Open subspaces of locally compact Hausdorff spaces are locally compact, but since I am not familiar with the notion of regularity I am looking for another explanation for the (apparent) fact that:
for any $x \in X-A$, I can find an open $V$ such that $x \in V$ and $\overline{V} \subset X-A$.
My efforts so far: Fix $x \in X - A$, then since $X$ is locally compact, $x$ has a compact neighborhood $B\subset X$. So there is a $U$ open in $X$ such that $x \in U \subset B$. $A$ is closed, so $X-A$ is open, therefore $U \cap (X-A)$ is open. This way, I was trying to construct a compact neighborhood of $x$ in $X-A$, but I can't figure out how. Any help is much appreciated!