Theorem 2.7: Suppose $U$ is open in a locally compact Hausdorff space $X$, $K \subset U$, and $K$ is compact. Then there is an open set $V$ with compact closure such that$$K \subset V \subset \overline{V} \subset U.$$
The given proof goes something like this. Because $K$ is compact it has finitely many open sets covering it. Let $G$ be the union of such sets, then $G$ is open and has a compact closure.
Let $C$ be the complement of $U$. By theorem 2.5, to each $p \in C$ there exists an open set $W_p$ that contains $K$ and $p \notin \overline{W_p}$. Thus $\{ C \cap \overline{G} \cap \overline{W_p} \}$, where $p$ ranges over $C$, is a collection of compact sets with an empty intersection. Then by theorem 2.6 there are points $p_1, \ldots, p_n \in C$ such that $$C \cap \overline{G} \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} = \emptyset.$$ Then we define $V = G \cap W_{p_1} \cap \ldots \cap W_{p_n}$ and we are done, $V$ satisfies the conclusion of the theorem.
My question is, why do we need $G$? The sets in the collection $\{ C \cap \overline{W_p} \}$ are also compact. ($\overline{W_p}$ is compact because $X$ is locally compact, and $C \cap \overline{W_p}$ is compact because $X$ is Hausdorff, and $C$ is closed.) Furthermore the intersection of the collection is empty and thus $$\hspace{-2in} (1)\hspace{2in}C \cap \overline{W_{p_1}} \cap \ldots \cap \overline{W_{p_n}} = \emptyset $$ for some $p_1, \ldots, p_n \in C$. If we define $V = W_{p_1} \cap \ldots \cap W_{p_n}$, then does't $V$ satisfy the conclusion of the theorem? By $(1)$, $\overline{V} \subset U$. Does my proof hold or am I missing something?
It is not necessarily true that the sets $\operatorname{cl}W_p$ are compact. Local compactness ensures that each point of $X$ has an open nbhd with compact closure, but each $W_p$ is an open nbhd of $K$, not of a point. Each $C\cap\operatorname{cl}G\cap\operatorname{cl}W_p$, however, is a closed subset of the compact set $\operatorname{cl}G$ and is therefore compact.
It’s actually in getting $G$ that we use local compactness. Each $x\in K$ has an open nbhd $H_x$ with compact closure and an open nbhd $G_x$ whose closure is contained in $U$; let $N_x=H_x\cap G_x$, and observe that $\operatorname{cl}N_x$ is compact and contained in $U$. $K$ is compact, so some finite $\mathscr{N}_0\subseteq\{N_x:x\in K\}$ covers $K$, and we take its union to be $V$: the fact that $\mathscr{N}_0$ is finite ensures both that $\operatorname{cl}V=\bigcup\{\operatorname{cl}N:N\in\mathscr{N}_0\}$ and that this union of compact sets is compact.