In a locally compact Hausdorff space, why are open subsets locally compact?

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!

Solution 1:

Suppose $U$ is an open subset of a locally compact Hausdorff space $X$, $K\subset U$ and $K$ is compact. Then there is an open set $V$ with compact closure such that $$K\subset V\subset \bar{V}\subset U.$$

This is Theorem 2.7 in Rudin's Real and Complex analysis. To prove it first you need to show (the easy) fact that in a Hausdorff space compact subsets and points can be separated. More precisely, if $K\subset X$ is compact and $p\in X\setminus K$, then there exist disjoint open sets $V, W$ containing $K$ and $p$, respectively.

For your purposes you need a special case of this theorem where $K$ is a point in the open set $U$.