A compact Hausdorff space is metrizable if it is locally metrizable 
A space $X$ is locally metrizable if each point $x$ of $X$ has a neighborhood that is metrizable in the subspace topology. Show that a compact Hausdorff space $X$ is metrizable if it is locally metrizable.
Hint: Show that $X$ is a finite union of open subspaces, each of which has a countable basis.

I tried to use the fact of compact space. But I do not know if the opens are compact subspaces.

[Hint: Using normality/regularity]
The space $X$ is compact Hausdorff, so it is normal/regular.
Now for each $x\in X$, there is a metrizable  open neighborhood $U_{x}$ of $x$, for this $U_{x}$, by regularity, there is an open neighborhood $V_{x}$ of $x$ such that
$x\in V_x \subset V_{x}^{-} \subset U_{x}$. Notice that the closure $V_x^{-}$ of $V_x$ is a closed subset of a compact space and a subspace of a metrizable space, therefore it is a compact metrizable space and has a countable basis, so $V_x$ has a countable basis. 
Since $X$ is compact, and the family $\{V_x\}_{x\in X}$ of opens covers $X$, there is a finite subcover in which every member is open and has a countable basis. Hence $X$ has a countable basis.
