Never Used Compact Closure... 
If $X$ is locally compact Hausdorff, then $X$ is regular.
Proof: Suppose that $x$ is some point not in the closed set $C$. Then $x$ is in the open set $C^c$, and $X$ being LCH implies there exists an open set $U$ such that $x \in U \subseteq \overline {U} \subseteq C^c$, where $\overline{U}$ is compact. Taking the complement, we get $C \subseteq \overline{U}^c$, and obviously $U \cap \overline{U}^c = \emptyset$.

Something about this seems fishy, especially since I never used the fact that $U$ has compact closure.
For completeness, we know:

Theorem 29.2: Let $X$ be Hausdorff. 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 $\overline V\subseteq U.$


Given that you know Theorem 29.2, then what you are using for local compactness is just that $U$ exists such that $\overline{U}\subseteq C^c.$
While you aren't using that $\overline U$ is compact, it is not true in general that $U$ exists if you don't have local compactness of $X.$
Indeed, "regular" can be seen as equivalent to:

Given any $x\in X$ and any neighborhood $V$ of $x$, there is a neighborhood $U$ of $X$ such that $\overline{U}\subseteq V.$

So Theorem 29.2 is stating that, for Hausdorff spaces, local compactness is a stronger form of this definition, with the added requirement that $\overline{U}$ is compact. The proof of 29.2 contains all the hard work.

(Correction: Contrary to my claim in prior versions of this document, this can all proceed "at $x$," rather than requiring global properties. Here's the outline.)
Interestingly, if we define "Hausdorff," "locally compact," and "regular" at a point as:

$X$ is "locally compact at $x$" if there is some open neighborhood $U$ of $x$ such that $\overline{U}$ is compact.

and:

$X$ is "regular at $x$" if, for any closed set $C$ not containing $x$, there exists disjoint open sets $U,V$ such that $x\in U, C\subseteq V.$

and:

$X$ is "Hausdorff at $x$" if, for any $y\neq x$ there are disjoint open sets $U,V$ with $x\in U$ and $y\in V.$

Then we get the following version of $29.2$:

Theorem 29.2-L: Given that $X$ is Hausdorff at $x$, then $X$ is locally compact at $x$ if and only if for any open neighborhood $V$ of $x$, we have an open neighborhood $U$ of $x$ such that $\overline{U}$ is compact and $\overline{U}\subseteq V.$
Proof: Let $W$ be the neighborhood of $x$ such that $\overline{W}$ is compact.
Let $U_1=W\cap V.$ Let $C=\overline{U_1}\cap V^c$, which is a closed subset of the compact $\overline U$ and hence compact.
For each $y\in C$, find a neighborhood $W_y$ of $y$ such that $x\not\in\overline{W_y}.$ (You can find such $W_y$ because $X$ is Hausdorff at $x.$)
Then $\{W_y\}_{y\in C}$ is an open cover of $C$, so there are $y_1,\dots, y_n$ such that $W_{y_1}\cup\cdots\cup W_{y_n}$ covers $C$.
Now let $U=U_1\setminus\left(\overline{W_{y_1}}\cup\cdots\cup\overline{W_{y_n}}\right).$
We know that $U$ is open and contains $x$ and is a subset of $W$, so $\overline{U}$ must be compact.
The last step is to prove that $\overline{U}\subseteq V.$ But $\overline{U}$ is contained in the closed set $\overline{U_1}\setminus\left(W_{y_1}\cup\cdots\cup W_{y_n}\right)$ which can't contain any elements of $V^c$, since any element in $\overline{U_1}\cap V^c=C$ is necessarily in some $W_{y_i}.$

