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False. We construct a compact Hausdorff space with a countable dense discrete subspace but no compatible metric. Letfor 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 $S=\{n_2,n_4,n_6,\dots \}$, 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.