I'd appreciate verification of this proof's validity (or lack thereof). The only slight concern I have is in the second to last sentence. This actually seems to be a nice example of the sort of thing I was looking for in this thread
"Let $X$ be a locally compact Hausdorff space, $A$ a closed subset of $X$, and $p$ a point not in $A$. Prove that there are disjoint open sets $U$ and $V$ in $X$ such that $p \in U$ and $A \subset V$."
Let $X_\infty$ denote the one-point compactification of $X$. Since $A$ is a closed subset of this compact space, it is compact itself in $X_\infty$. Since $X$ is Hausdorff and locally compact, $X_\infty$ is Hausdorff and so there are disjoint open sets in $X_\infty$ containing the disjoint compact subsets $\{p\}$ and $A$ by Theorem 6.5*. Since the open sets of $X$ are open considered as subsets of $X_\infty$, these disjoint open sets can come from $X$. Therefore, there are disjoint open sets $U$ and $V$ in $X$ with $p \in U$ and $A \subset V$.
*Theorem 6.5: "If $A$ and $B$ are disjoint compact subsets of a Hausdorff space $X$, then there exist disjoint open sets $U$ and $V$ in $X$ such that $A \subset U$ and $B \subset V$."
Best Answer
$\newcommand{\cl}{\operatorname{cl}}$The argument can be made to work, but not quite as you’ve stated it. The problem is that you have no reason to think that the open sets $U$ and $V$ given you by Theorem 6.5 do lie in $X$, so one further step is necessary: the definition of $X_\infty$ ensures that $U\cap X$ and $V\cap X$ are open in $X$, and of course $(U\cap X)\cap(V\cap X)=U\cap V\cap X=\varnothing$, $p\in U\cap X$, and $A\subseteq V\cap X$. Thus, $U\cap X$ and $V\cap X$ give you the desired separation in $X$.
It’s by no means necessary to invoke $X_\infty$, howver, to prove that $X$ is $T_3$. Just let $U=X\setminus A$, let $V$ be an open nbhd of $p$ with compact closure, and let $W=U\cap V$. Then $\cl W$ is compact and Hausdorff and therefore $T_3$, and $W$ is open in $\cl W$, so there is a set $G$ open in $\cl W$ such that $$p\in G\subseteq\cl G\subseteq W\;.$$ It’s easy to verify that $G$ is actually open in $X$ (why?), so $G$ and $X\setminus\cl G$ form the desired separation.