= Solution
\b[False.] We construct a <compact Hausdorff space> with a countable dense discrete subspace but no compatible <metric>. Let
$$
P=\{0,1\}^{\mathcal P(\mathbb N)},\qquad
j(n)_S=\begin{cases}1&n\in S,\\0&n\notin S,\end{cases}
$$
for each subset $S\subseteq\mathbb N$. Give $P$ its <product topology>, and let $X=\overline{j(\mathbb N)}$ in $P$. The <Tychonoff theorem> makes $P$ compact, and the product is Hausdorff; hence $X$ is a <compact Hausdorff space>. By construction, $Y=j(\mathbb N)$ is countable and dense in $X$, so $X$ is a <separable topological space>. The coordinate $S=\{n\}$ isolates $j(n)$ within $Y$. Thus the subspace topology on $Y$ is discrete and is induced by the <discrete metric>.
Suppose $X$ were metrizable. An infinite sequence of distinct points of $Y$ would have a convergent subsequence, by sequential compactness of a <compact metric space>. Write this subsequence as $j(n_k)$ with distinct $n_k$. For $S=\{n_2,n_4,n_6,\ldots\}$, its $S$ coordinate alternates between zero and one and therefore does not converge. But coordinate projections are continuous in the <product topology>, so every coordinate of a convergent sequence must converge. This contradiction proves that $X$ is not metrizable. This is a <separable compactification need not be metrizable> counterexample.
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