TITLE: Compact Hausdorff spaces without isolated points in ZF
QUESTION [8 upvotes]: S is uncountable :=  |$\mathbb{N}$| < |S|

S is noncountable :=  |S| $\not\leq |\mathbb{N}|$

(X,$T$) is a nice space :=  (X,$T$) is a compact Hausdorff space without isolated points
Does [ ZF / ZF + Countable Choice ] prove that every nice space is [ uncountable / noncountable ] ?

If not, is it known to prove that the statement implies some choice principle?

What if the spaces are additionally assumed to be metrizable?
Now, that's basically 12 questions, so I certainly don't anticipate answers for all of them.

If it matters, the one I'm most curious about is "Does ZF prove that every nice space is noncountable?".

REPLY [5 votes]: I'd like to extend on Joel's answer, and point that ZF itself cannot prove that every "nice space" is uncountable.
It is consistent that the axiom of choice fails and there exists a Dedekind-finite set $X$ which can be topologized as follows:

$X$ is Hausdorff compact;
$X$ is strongly connected (every real valued function is constant);
The topology is an order topology of a dense linear order with endpoints (if we remove the endpoints then this is a locally-compact space, and every closed interval is compact).

It follows that there are no isolated points, so this is a nice space. However the set itself is Dedekind-finite and so not uncountable (but it is still noncountable). Such topological space is called a Läuchli continuum.
For example if we consider Mostowski's ordered model, by adding two endpoints to the atoms the result is a Dedekind-finite $X$ with the above properties.
I elaborated on this in a recent math.SE answer which includes all the relevant references and more.

It should be pointed that in the great book of choice principles the existence of non-trivial Läuchli continua is the negation of Form 155; whereas countable choice is Form 8. I could not find much connection between the two forms in the site.