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?