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This is exercise 7 from section 34 in Munkres. The hint given is to show that the space is a union of finitely many subspaces which are second countable. This question has been asked before A compact Hausdorff space is metrizable if it is locally metrizable.

However, I'm still confused. The idea appears to be: let $U_x$ be a metrizable neighborhood of $x$ for each $x\in X$. Since $X$ is compact, there are $x_1,\ldots,x_n$ for which $\{U_{x_k}\}$ form a finite subcover. If we knew that each $U_{n_k}$ was second countable, we could conclude that $X$ was metrizable. This is because $X$ is regular and second countable.

I can't find any reason why the $U_{n_k}$ are second countable though.

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  • $\begingroup$ @Mirko Why do we know that each $U_{x_k}$ is second countable? $\endgroup$ – Tim Raczkowski Dec 12 '15 at 23:43
  • $\begingroup$ It's also true for paracompact Hausdorff spaces, though the proof is a bit harder. $\endgroup$ – Henno Brandsma Dec 13 '15 at 7:14
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Choose $U_x$ compact ( with interiors covering $X$) and metrizable. Then they are second countable. ( a compact metric space is second countable).

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  • $\begingroup$ The $U_x$ are open. $\endgroup$ – Tim Raczkowski Dec 12 '15 at 23:44
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    $\begingroup$ @Tim Raczkowski: Inside every open metrizable open neighborhood of $x$ choose a compact subset containing $x$ in the interior. That one will have a countable basis. Finitely many of these will cover $x$. The interior of these sets will have a countable basis. So $X$ will have a countable basis. $\endgroup$ – Orest Bucicovschi Dec 12 '15 at 23:50
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    $\begingroup$ @Tim Raczkowski: The fact used is that for every point $x$ in an open subset $U$ of a compact topological space there exists $K$ compact so that $x \in \overset{\circ}{K }$ and $K \subset U$. $\endgroup$ – Orest Bucicovschi Dec 12 '15 at 23:53
  • $\begingroup$ Right. Let's say that for each $U_{x_k}$ we have a $K_{x_k}$. It's not clear to me that that the $K_{x_k}$ cover $X$. Most likely $K_{n_k}\subsetneq U_{n_k}$. $\endgroup$ – Tim Raczkowski Dec 12 '15 at 23:56
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    $\begingroup$ @TimRaczkowski : Interior of a set $\endgroup$ – Orest Bucicovschi Dec 13 '15 at 0:05
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For each $p\in X$ let $ U_p$ be a ndhd of $ p$ such that $U_p$ is metrizable. Since $X$ is a compact Hausdorff space, it is a regular space, so there is an open $V_p$ such that $p\in \overline {V_p} \subset U_p.$ Since $X$ is a compact Hausdorff space, $\overline { V_p}$ is compact. And $\overline { V_p}$ is metrizable because it is a subset of $U_p.$ Now $\cup \{V_p :p\in X\}=X$, so $\cup \{V_p :p\in F\}=X$ for some finite $F$ because $X$ is compact. Each $\overline {V_p}$ is second -countable because it is metrizable and compact. Any subspace of a second-countable space is second-countable. So each $V_p$ is second countable, and is open, and $F$ is finite, so $X=\cup \{V_p:p\in F\}$ is second-countable.

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This answer takes some ideas from Orest Bucicovschi's answer and comments and puts them together + adds more detail.

First, we need 4 claims which I will not prove here (can be found in Munkres etc'):

  1. Every compact Hausdorff space is regular
  2. In a regular space $X$, for every point $x\in X$ and every open set $U$ containing it, there exists another open set $V$ such that $x\in V\subseteq\overline{V}\subseteq U$
  3. Every closed subset of a compact space is compact
  4. Every compact metric space is second countable

Using the first three claims, we can prove that for every point $x$ in an open subset $U$ of a compact topological space, there exists a compact set $K$ so that $x\in \mathring{K}$ (interior of $K$) and $K\subseteq U$. This compact set $K$ will be exactly $\overline{V}$ from the second claim mentioned above.

Then, if we take an open cover of $X$ using the interiors of compact sets $K_x$, it has a finite sub-cover, and therefore we can build a finite cover of $X$ using compact metrizable sets $K_x$. According to the 4th claim, each of these sets has a countable basis, and we can show that the union of these bases is a countable basis for all of $X$. Therefore, $X$ is second countable, and from there we can continue as you mentioned in your question.

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