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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.

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    Let $x$ be any point, and $U$ an open neighbourhood of $x$. Since $U$ is locally compact, $x$ has a compact neighbourhood in $U$. What is the relation between the neighbourhoods of $x$ in $U$ and the neighbourhoods of $x$ in the whole space? – 2017-02-05
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    What is your exact definition of $X$ being locally compact? Every point has a compact neighbourhood? This will matter for the proof. In your case a local base consists of neighbourhoods of the point ( having that point in its interior), not necessarily open sets? – 2017-02-05
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    @HennoBrandsma. Yes, exactly. This are the definitions I am using. – 2017-02-05
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    Are you allowed (at this stage) to use that $X$ is a regular space? – 2017-02-05
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    @HennoBrandsma. Yes, that is fine. As I wrote in my question, I have in fact a proof using the regularity of the space, but as far as I can see it does not follow the outline of the proof given in the exercise. – 2017-02-05
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    So you are not asking how to prove the first fact (open subspace?) only how the second (local base) follows from the first? – 2017-02-05
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    @HennoBrandsma. Yes. – 2017-02-05
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53077/discussion-between-henno-brandsma-and-jvnv). – 2017-02-05

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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 $x$.