Here is Prob. 10, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ is a Hausdorff space that is locally compact at the point $x$, then for each neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.
My Attempt:
Suppose that $X$ is a Hausdorff space, that $x \in X$, and that $X$ is locally compact at the point $x$. Let $U$ be any neighborhood of $x$ in $X$ (i.e. an open set containing $x$).
Since $X$ is locally compact at $x$, there exists a compact subspace $C$ of $X$ and a neighborhood $U^\prime$ of $x$ such that
$$ U^\prime \subset C. \tag{0} $$
Now let us put
$$ V \colon= U \cap U^\prime. \tag{Definition 0} $$
Then $V$ is a neighborhood of $x$, and (by (Definition 0) and (0) above), we obtain
$$ V \subset U^\prime \subset C, $$
and hence
$$ V \subset C. \tag{1} $$
Now as $C$ is a compact subspace of the Hausdorff space $X$, so $C$ is closed in $X$, by Theorem 26.3 in Munkres.
Moreover, as $C$ is a closed set in $X$ and as $V \subset C$ by (1) above, so we can also conclude that
$$ \overline{V} \subset C. \tag{2} $$
And, as $\overline{V}$ is a closed set in $X$ and as $\overline{V} \subset C$, so
$$ \overline{V} = \overline{V} \cap C, $$
and thus by Theorem 17.2 in Munkres we can also conclude that $\overline{V}$ is also closed in the subspace $C$ of $X$
Finally, as $\overline{V}$ is a closed set in the compact space $C$, so $\overline{V}$ is also compact, by Theorem 26.2 in Munkres; as $\overline{V}$ is compact as a subspace of $C$ and as $C$ is a subspace of $X$, so $\overline{V}$ is also compact as a subspace of $X$.
What next? How to proceed from here to show that $\overline{V} \subset U$ also?
Best Answer
You need to use that $C$, being compact in a Hausdorff space, is (a normal and hence) regular space, and in a regular space we have, whenever $x \in O$ open, an open $O'$ such that $x \in O' \subseteq \overline{O'} \subseteq O$.
Apply this to your $U \cap U'$ and inside we get the required $V$ almost for free from this regularity (open in open is open again, and $\overline{V}$ sits inside $C$ so is compact).
Added upon request:
Without using regularity directly, in Lemma 26.4 Munkres shows that whenever $Y$ is a compact subset of a Hausdorff space $X$ and $x_0 \notin Y$, there are open and disjoint $U$ and $V$ containing $x_0$ resp. $Y$, which is all the regularity we need:
We have $C$ compact and $U'$ the open neighbourhood of $x$ inside $C$. Then $C':=C \cap \left( X \setminus \left( U \cap U^\prime \right) \right)$ is compact (being closed in $C$) and $x \notin C'$. So by 26.4 we have open $x \in O_1$ and $O_2 \ \supseteq C'$ that are disjoint.
It's then not too hard to see that $V= U \cap U' \cap O_1$ is as required, it's certainly an open neighbourhood of $x$ and $\overline{V} \subseteq U$.
(For the last inclusion, which is clear (for me at least) from a picture, a more formal proof:
Suppose $y \in \overline{V}$, then in particular $y \in \overline{U'} \subseteq \overline{C}=C$ (using Hausdorff again!; also when we applied 26.4 of course). Also $y \in \overline{O_1}$, so $y \notin O_2$ (any point in $O_2$ has a neighbourhood, namely $O_2$, that is disjoint from $O_1$ to guarantee that it's not in the closure of $O_1$), and so $y \notin C'$ either. Knowing the definition of $C'$ and $y \in C$ we see that $y \notin X\setminus (U \cap U')$ which implies $y \in U' \cap U \subseteq U$ as required.
The regularity route is easier, so that's why I chose it first, but Munkres treats regularity after compactness, hence this addition.