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I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number (or something equivalent), not Dedekind-finite. Obviously, the empty set's only topology is compact Hausdorff.

Suppose $X$ is a non-empty set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite. Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology. Moreover, $\mathcal T$ is discrete if and only if $X$ is finite.

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number (or something equivalent), not Dedekind-finite. Obviously, the empty set's only topology is compact Hausdorff.

Suppose $X$ is a non-empty set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite. Then $\mathcal T$ is a compact Hausdorff topology on $X.$ In particular, if $X$ is infinite, then $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology; if $X$ is finite, then $\mathcal T$ is discrete.

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number, not Dedekind-finite. Obviously, finite sets can be given the discrete topology. Suppose $X$ is an infinite set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite.

Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology.

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number (or something equivalent), not Dedekind-finite. Obviously, the empty set's only topology is compact Hausdorff. 

Suppose $X$ is a non-empty set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite. Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology. Moreover, $\mathcal T$ is discrete if and only if $X$ is finite.

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number, not Dedekind-finite. Obviously, finite sets can be given the discrete topology. Suppose $X$ is an infinite set, fix $x\in X$, and let $Y:=X\smallsetminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite.

Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology.

I see that I'm rather late to the party. Here's an answer to the following question that you asked in the comments above:

"[I]s it conceivable that it is a weak AC principle that every set has a compact Hausdorff topology?"

In fact, there is no need for any choice principle at all, if by finite, we mean in bijection with a natural number, not Dedekind-finite. Obviously, finite sets can be given the discrete topology. Suppose $X$ is an infinite set, fix $x\in X$, and let $Y:=X\setminus\{x\}.$ Now let $\mathcal T$ be the set of all subsets $U$ of $X$ such that either (1) $U\subseteq Y$ or (2) $x\in U$ and $X\setminus U$ is finite.

Then $\mathcal T$ is a compact Hausdorff topology on $X$, and in particular, $\langle X,\mathcal T\rangle$ is homeomorphic to the Alexandrov one-point compactification of $Y$ in the discrete topology.