Theorem 29.2. Let $X$ be a Hausdorff space. 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 $\bar{V}$ is compact and $\bar{V} \subseteq U$.
Proof: Clearly this new formulation implies local compactness; the set $C = \bar{V}$ is the desired compact set containing a neighborhood of $x$. To prove the converse, suppose $X$ is locally compact, let $x$ be a point of $X$ and let $U$ be a neighborhood of $x$. Take the one-point compactification $Y$ of $X$, and let $C$ be the set $Y - U$. Then $C$ is closed in $Y$, so that $C$ is a compact subspace of $Y$. Apply Lemma 26.4 to choose disjoint open sets $V$ and $W$ containing $x$ and $C$, respectively. Then the closure $\bar{V}$ of $V$ in $Y$ is compact, furthermore, $\bar{V}$ is disjoint from $C$, so that $\bar{V} \subset U$, as desired.
So, in the theorem above it is equivalent to the definition of locally compactness, but what I don't understand in the proof above is that it holds that the desired $V$ works, but only for the one point compactification space $Y$ in the theorem above, so why does it also work for the space $X$ itself ?
It’s a general fact that if $X$ is a subspace of a space $Y$, and $K$ is a subset of $X$ that is compact in $Y$, then $K$ is compact in $X$ as well. If you’ve not seen this before, you should try to prove it. Start with a cover $\mathscr{U}$ of $K$ by sets that are open in $X$. For each $U\in\mathscr{U}$ there is a $V_U$ open in $Y$ such that $U=X\cap V_U$. Let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$; then $\mathscr{V}$ is a cover of $K$ by sets that are open in $Y$, so it has a finite subcover. Can you finish the argument from there?
In this particular setting let $K=\operatorname{cl}_YV$. $K$ is certainly compact in $Y$. Moreover, $K\cap W=\varnothing$, so $K\subseteq Y\setminus W\subseteq Y\setminus C\subseteq X$. The general fact now applies, and we conclude that $K$ is compact in $X$.