 Separable compactification need not be metrizable

Embed $\mathbb{N}$ into the compact product $\{0,1{\}}^{\mathcal{P}(\mathbb{N})}$ by $j(n{)}_S=1_S(n)$ and take its closure $X$. The Tychonoff theorem makes $X$ a compact Hausdorff space, while $j(\mathbb{N})$ is countable, dense and discrete in its subspace topology. Yet $X$ is not metrizable. Any convergent subsequence $j(n_k)$ of distinct points would have a convergent coordinate for $S=\{n_2,n_4,\dots \}$, whereas that coordinate alternates between zero and one. This contradicts sequential compactness of a compact metric space. Thus a dense metrizable subspace does not force a separable compact Hausdorff ambient space to be metrizable.