A Compact Hausdorff space which is locally metrizable is metrizable. 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.
 A: Choose $U_x$ compact ( with interiors covering $X$) and metrizable. Then they are second countable. ( a compact metric space is second countable). 
A: 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.
A: 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'):


*

*Every compact Hausdorff space is regular

*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 $x\in V\subseteq\overline{V}\subseteq U$

*Every closed subset of a compact space is compact

*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.
