= Countable product compactness implies countable choice
This concerns arbitrary <compact spaces>, with no <Hausdorff> restriction. For nonempty sets $A_j$, adjoin a distinguished point $\infty_j$ and use the four-open-set <topology> $\{\varnothing,A_j,\{\infty_j\},A_j\cup\{\infty_j\}\}$. Each resulting space is compact without choice. The product has the canonical all-infinity point. The closed coordinate conditions $x_j\in A_j$ have the <finite intersection property>, since finite choice is provable without the <axiom of choice>. If the countable product is compact, their intersection is nonempty, and any point in it supplies a <choice function>.
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