Local compactness exercise Let $X$ be a Hausdorff locally compact in $x \in X$. Show that for each open nbd $U$ of $x$ there exists an open nbd $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subset U$.
My work:
Since $X$ is Hausdorff and locally compact then $X$ is regular. Let $U$ be an open nbd of $x$. By assumption $X$ is locally compact so there exists some open nbd $W$ of $x$ such that $\overline{W}$ is compact. Now consider the open set $W \cap U$ this is non-empty since $x$ lies in the intersection. By regularity find an open set $V$ such that:
$x\in V \subset \overline{V} \subset W \cap U$
Then in particular $\overline{V} \subset U$. But also $\overline{V} \subset W \subset \overline{W}$. Since $\overline{W}$ is compact then $\overline{V}$ is a closed subset of a compact set, hence compact.
Is the above OK? Thank you.
 A: Since I have already commented regarding the validity of your proof, let me mention a couple of interesting relevant exercises:
Exercise 1: If $X$ is a locally compact Hausdorff space, then prove that $X$ is completely regular, that is, prove that for every point $x\in X$ and every closed set $C$ with $x\not\in C$, there exists a continuous function $F:X\to [0,1]$ such that $F(x)=0$ and $F(C)=\{1\}$. (Hint: examine the proof of Urysohn's lemma carefully.)
Exercise 2: If $X$ is a locally compact Hausdorff space and $\{\infty\}$ is a singleton set disjoint from $X$, then define $Y=X\cup \{\infty\}$. Furthermore, define a topology on $Y$ by dictating the following subsets of $Y$ to be open:
(1) If $U\subseteq X$ is open in $X$, then $U$ is open in $Y$.
(2) If $V=U\cup \{\infty\}$ where $U$ is open in $X$ and $X\setminus U$ is a compact subset of $X$, then $V$ is open in $Y$.
(a) Prove that $Y$ is a topological space.
(b) Prove that $Y$ is compact Hausdorff. (Hint: to prove that $Y$ is Hausdorff, you will need local compactness and to prove that $Y$ is compact, you will need to think about the role played by $\infty$ in the topological space $Y$.)
(c) If $X=\mathbb{N}$ with the discrete topology, then prove that $Y$ is 
homeomorphic to $\{0\}\cup \{\frac{1}{n}:n\in\mathbb{N}\}$. 
(d) If $X=\mathbb{R}^n$ with the standard topology, then prove that $Y$ is homeomorphic to the sphere $S^n\subseteq \mathbb{R}^{n+1}$.
I hope these exercises are useful!
A: I'll just point out that using the fact that $X$ is regular is a bit of overkill.  Here's a simpler fact, which is adequate here:

Lemma.  If $X$ is Hausdorff, $x \in X$, and $K \subset X$ is compact, then there exist disjoint open neighborhoods of $x$ and $K$.
Proof.  Exercise.

Now, for your problem, assume without loss of generality that $U$ is precompact.  (Otherwise, replace $U$ by its intersection with a precompact neighborhood of $x$.)  Then the boundary $\partial U$ is compact.  Let $V,W$ be disjoint open neighborhoods of $x$, $\partial U$.  It's easy to verify that $V \cap U$ has the desired property.  (Drawing a picture may help.)
A: Solution. Let $U$ be an arbitrary nbd of $x$. By definition, there
exist compact set $C\subset X$ with a nbd $W$ of $x$ such that $W\subset C$.
Now put $Y=C\setminus \left( U\cap W\right) $. Since $C$ is closed ($X$ is
Hausdorff), then $Y=C\setminus U\cap W$ must be also closed subset of $C$.
Hence $Y$ is compact. Furthermore, $x\notin Y$. Due to lemma cited above,
there are two disjoint open sets $O_{1}$ and $O_{2}$ such that $x\in O_{1}$, 
$Y\subset O_{2}$. Now if we put $V=O_{1}\cap U\cap W$, then $V$ suffices the
desired properties.$\Box $
