Let $X$ be a Hausdorff locally compact in $x \in X$. Show that for each open nbd $U$ of $x$ there exists an open nbd $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.
My work:
Since $X$ is Hausdorff and locally compact then $X$ is regular. Let $U$ be an open nbd of $x$. By assumption $X$ is locally compact so there exists some open nbd $W$ of $x$ such that $\overline{W}$ is compact. Now consider the open set $W \cap U$ this is non-empty since $x$ lies in the intersection. By regularity find an open set $V$ such that:
$x\in V \subset \overline{V} \subset W \cap U$
Then in particular $\overline{V} \subset U$. But also $\overline{V} \subset W \subset \overline{W}$. Since $\overline{W}$ is compact then $\overline{V}$ is a closed subset of a compact set, hence compact.
Is the above OK? Thank you.