The main theorem is as follows. I think most people are familiar with that:
Theorem. Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if for every $x\in X$ and every open set $U$ containing $x$, there exists a neighborhood $V$ of $x$ such that ${\rm Cl}(V)$ is compact and ${\rm Cl}(V)\subseteq U$.
One direction is trivial, so we only need to show that the condition holds if $X$ is locally compact.
In my definition:
Definition. A topological space $X$ is locally compact if for every $x\in X$, there is a compact subset $C$ of $X$ such that $x\in{\rm Int}(C)$.
I know there are many proofs available to that theorem, but I wonder if I can prove it without referring to the one-point compactification. Here follows my proof, which uses the regularity of locally compact Hausdorff space.
Proof. Suppose $X$ is locally compact. For each $x\in X$, let $C$ be a compact subset of $X$ with $x\in{\rm Int}(C)$. For every neighborhood $U$ of $x$, since $X$ is regular, there exists a neighborhood $V'$ of $x$ such that ${\rm Cl}(V')\subseteq U$. Then we set \begin{equation*} V=V'\cap{\rm Int}(C). \end{equation*} Apparently, $V$ is a neighborhood of $x$ where \begin{equation*} {\rm Cl}(V)={\rm Cl}(V'\cap{\rm Int}(C))\subseteq{\rm Cl}(V')\cap{\rm Cl}({\rm Int}(C))\subseteq{\rm Cl}(V')\cap C. \end{equation*} On the one hand, we have ${\rm Cl}(V)\subseteq{\rm Cl}(V')\subseteq U$. On the other hand, since ${\rm Cl}(V)$ is closed in $C$ and $C$ is compact, we can see that ${\rm Cl}(V)$ is also compact, as desired.
If anyone finds it interesting, could you please help me check whether my proof is valid? Any help will be appreciated.
I think it's better not to rely on $X$ being (completely) regular (which is also most easily proved by using the one-point compactification) but by using the classic fact that a compact Hausdorff space is normal (and hence regular).
So if $x \in O \subseteq C$ with $O$ open and $C$ compact (as the assumption of local compactness gives us) and $U$ is any open set containing $x$, then $U \cap O$ is open in $C$ which is (as said) regular and so we find an open neighbourhood $V$ of $x$ (open in $C$, so of the form $V=V' \cap C$ for some $V'$ open in $X$) such that $\operatorname{cl}_C(V) \subseteq U\cap O$ and then check that $V' \cap O$ is as required.