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I came across a theorem in Munkres' Topology.

Theorem. Let $X$ be a Hausdorff space $X$ is locally compact iff given $x \in X$ and given a neighbourhood $U$ of $x$ there is a neighbourhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.

Proof in the book is as follows: For the $(\implies)$ part, suppose that $X$ is locally compact, let $x \in X$ and $U$ is a neighbourhood of $x$. Take the one-point compactification $Y$ of $X$. (This $Y$ is compact and Hausdorff, closure of $X$ is $Y$ and $Y-X$ is a singleton.) Define $C = Y-U $. Since $U$ is open, $C$ is closed in $Y$. $Y$ is compact so that $C$ is also compact since it is closed. Then by using the lemma below:

Lemma: $X$ is a Hausdorff space, $Y$, a compact subspace of $X$ and for a point $x_0 \in X$ suppose that $Y$ does not contain $x_0$. Then, there exist 2 open subsets of $X$, $U$ and $V$ such that $U$ contains $Y$ and $V$ contains $x_0$.

We can find open and disjoint subsets $V$ and $W$ of $Y$ such that $V$ contains $x$ and $W$ contains the set $C$. Then, since $\overline{V}$ is closed, it is a compact subspace of $Y$ and it is disjoint from $C$. So, we found a compact subset $\overline{V}$ such that $\overline{V} \subset U$.

The other side of the theorem can be proved by setting $C=\overline{V}$.

Proof of this theorem seems very elegant to me, however, the construction is not obvious for me. How can we prove it in a more "obvious" way? Also if you have any suggestions for the way that studying compactness in general, I appreciate that.

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