If $X$ is locally compact Hausdorff, then $X$ is regular. The following are from Munkres' book on topology:

Theorem 29.2. Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if given $x$ in $X$, and given a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$.
Lemma 31.1 a. Let $X$ be a topological space. Let one-point sets in $X$ be closed. $X$ is regular if and only if given a point $x$ of $X$ and a neighborhood $U$ of x, there is a neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.

Doesn't the $(\implies)$ part of theorem 29.2 and the $(\impliedby)$ part of lemma 31.1 a together prove that locally compact Hausdorff spaces are regular? I'm a little bit suspicious because this seems an easy solution, I might be missing something.

In fact, X is completely regular: Let X be locally compact Hausdorff. Then, there exists a compact Hausdorff space Y containing X as a subspace. Now since Y is normal, Y is completely regular, which implies that X is completely regular, because it is a subspace of a completely regular space.
