= Separable compactification need not be metrizable
Embed $\mathbb N$ into the compact product $\{0,1\}^{\mathcal P(\mathbb N)}$ by $j(n)_S=\mathbf1_S(n)$ 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 $S=\{n_2,n_4,\ldots\}$, 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.
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