TITLE: Hausdorff space in which each point has a compact neighbourhood is locally compact
QUESTION [5 upvotes]: Could you help me prove the following fact? I've been trying to prove it and I've searched for a hint in Englking's book, but I haven't come up with anything:
If $X$ is a Hausdorff space and each $x \in X$ has at least one compact neighbourhood, then $X$ is locally compact.
$U$ is a neighbourhood of $x \in X$ $\iff$ $\exists V$ - open $: x \in V \subset U$.
$X$ is locally compact $\iff$ each $x \in X$ has a basis of compact neighbourhoods.

REPLY [6 votes]: A tricky one, indeed. I already had that once. Here we go :
We suppose that every point of $X$ has at least one compact neighborhood. 
Let $x\in X$ and let $V$ be an open neighborhood of $x$. 
We want to show that there exists a compact neighborhood $W$ of $x$ with $W \subset V$.
By hypothesis, $x$ has a compact neighborhood, let's call it $C$.
Let's take the subspace topology for $C$, so we have that $C$ is a compact Hausdorff space. 
Let's now take  $U = C \cap V$. And now we have that in $C$, $U$ is an open neighborhood of $x$. We know that $C$ is locally compact since it's compact and Hausdorff. 
Thus, in $C$  we have a compact neighborhood $W$ of $x$ with $W \subset U$, and then also $W \subset V$. 
In the end, we must check that $W$ is a neighborhood of $x$ in $X$, too. But since a neighborhood in a neighborhood is a neighborhood, we have it directly.