If $X$ is a LCHS and $K, O \subseteq X$ with $K$ cpt & $O$ open, then $\exists U$ open s.t. $K \subseteq U \subseteq \overline{U} \subseteq O$?

Question

I'm having trouble fully understanding the proof of this statement.

Suppose $X$ is a locally compact Hausdorff topological space. Then if $K$ is a compact subset of $X$ and $O$ is any open subset of $X$ containing $K$, $\exists$ an open subset $U$ of $X$ such that $K \subseteq U \subseteq \overline{U} \text{ compact} \subseteq O$, with $\overline{U}$ being the closure of $U$.

Here is how I was taught the proof goes:

Let $K \subseteq X$ be compact, and let $O \subseteq X$ be open such that $K \subseteq O$. Then since $O$ is open, $O^{c}$ is closed. But $O^{c}$ closed and $K$ compact implies we can find $U, V$ open such that $K \subseteq U$ and $O^{c} \subseteq V$ with $U \cap V = \emptyset$ (this is because locally compact Hausdorff topological spaces are regular). But if $O^{c} \subseteq V$, then $V^{c} \subseteq O$.

Since $V$ is open, $V^{c}$ is closed. Also, $U \cap V = \emptyset \implies U \subseteq V^{c}$. Since $V^{c}$ is closed, we have the closure of $U$, $\overline{U}$, is contained in $V^{c}$. So, we have $K \subseteq U \subseteq \overline{U} \subseteq O$. We can assume $\overline{U}$ is compact, and so we are done.

First question: Is there an easy way to prove that a locally compact Hausdorff topological space is regular? I wasn't able to do so on my own.

Second question: Why can we assume $\overline{U}$ is compact at the end? Does it have anything to do with the space being locally compact?

Answer 1

First question: In this link, you can find your desired proof. It invokes the one point compactification and I think that it is a standard one.

I would like to note that, in your case, where you are trying to separate a compact set from a closed set, the proof can be done without invoking regularity (at least, it does not use it explicitly). For the proof, take a look in Theorem 2.7. of Rudin's book.

Second question: Yes, it is related with locally compactness. Indeed, once $K$ is compact, you can find a open neighbourhood of $K$ with compact closure. Indeed, for each $p\in K$, there is $p\in U_p$ open with $\overline{U}_p$ compact. The Family $\{U_p\}_{p\in K}$ is an open cover of $K$, thus, we can assume that $$K\subset \bigcup_{i=1}^N U_{p_i},$$

for some finite $N$, therefore, $$K\subset \bigcup_{i=1}^N \overline{U_{p_i}},$$

which is compact.