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.
 A: Of course, if $X$ is metrizable then it is also locally metrizable. 
For the other direction, you could really follow the hint. Supposed that $X$ is locally metrizable, there exists an open cover $\bigcup_{x\in X}U_x$ of $X$ where $U_x$ is a metrizable neighborhood of $x$. Because of compactness, it has a finite subcover: $\bigcup_{i<n} U_i=X$ where $U_i:=U_{x_i}$ for some $x_i$.
Because of metrizability, each $U_i$ has countable base $(V_{i,j})_j$ of open subsets, so that all finite intersections of these are still countable, and they give a base of $X$.
The space also has to satisfy the separation axiom $T_3$, but this holds as $X$ is Hausdorff and compact.
A: For every $x\in X$, there exists a neighborhood $U_x$ which is metrizable. These neighborhoods cover $X$, i.e., $X=\bigcup_x U_x$. Now use the definition of compactness to reduce this to a finite union, $X=U_1\cup\ldots\cup U_n$. Each of these sets is metrizable, so pick metrics which are defined locally on each $U_i$. Lastly, use a partition of unity to patch together the local metrics into a global one.
A: [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.
