I am trying to solve the following exercise:
Show that an open subset of a compact Hausdorff space is locally compact. Use this to conclude that any locally compact Hausdorff space has a local base consisting of compact sets.
Here, a topological space is locally compact if each point possesses a compact neighbourhood, and a local base is a collection such that for any given neighbourhood of a point there is a neighbourhood in the local base that is contained in the given neighbourhood.
It is clear to me why an open subset of a compact Hausdorff space is locally compact. However, I do not see how to use this to conclude that any locally compact Hausdorff space has a local base consisting of compact sets.
I am aware of another solution of the second part of the above stated exercise (using the complete regularity of the space), but I would also like a proof that follows the method outlined in the exercise.
Any help or comment is highly appreciated.
Suppose $X $ is open and $x \in O$ as $O$ is locally compact by the first fact it has a compact neighbourhood $C$ in $X$ so $x \in \operatorname{Int_O}(C)$ and $C$ compact (compactness is absolute). As $O$ is open, so is $\operatorname{Int_O}(C)$ so $C$ is still a neighbourhood of $x$ in $X$, and also still compact. So compact neighbourhoods form a local base at $x$ as we can do this for every open $O$ that contains $x$.