Show that a locally compact Hausdorff space is regular.

For the first approach, note that $X\in\tau_\infty$, which immediately yields $V_\infty\setminus\{\infty\}\in\tau$.

Suppose $X$ is a Hausdorff space which is locally compact, meaning that every point has a compact neighborhood. Consider a closed set $F\subseteq X$ and a point $x\in X\setminus F.$ Let $K$ be a compact neighborhood of $x.$

Since $X$ is Hausdorff, and since $F\cap K$ is compact and $x\notin F\cap K,$ there are disjoint open sets $U,V$ such that $x\in U$ and $F\cap K\subseteq V.$

Now $x\in\operatorname{int}K$ (since $K$ is a neighborhood of $x$), and $K$ is closed (since $K$ is compact and $X$ is Hausdorff).

Thus we have disjoint open sets $U_0=U\cap\operatorname{int}K$ and $V_0=V\cup(X\setminus K)$ with $x\in U_0$ and $F\subseteq V_0.$

The second approach can be made clearer (for my tastes, which do not have to agree with yours) by using the English language more.

A simple reformulation is that you are looking for a closed neighbourhood of $x$ that does not meet $F$. Now $x$ has a compact neighbourhood $K$ and a closed neighbourhood $W_K$ in $K$ that does not meet $F$. Since $K$ is a neighbourhood of $x$, $W_K$ is also a neighbourhood of $x$ in $X$, and since $K$ is closed, $W_K$ is also closed in $X$.

(The closed neighbourhood $W_K$ can be chosen as the closure of $U_K$ in your formulation of the proof.)