If every infinite subset has a limit point in a metric space $X$, then $X$ is separable (in ZF)
Yesterday, i posted this question and got an answer that 'Limit Point Compact⇒Separable' is unprovable in ZF.
As you can see, the proof in the link is done by the fact that there could be an infinite set which doesn't have a countable subset. I accepted the equivalence as an consequence of AC and continued my study.
However, most of spaces in classical analysis are Dedekind-infinite(i.e. $X≧|\aleph_0|$) as far as i know, and i have no idea how to prove this in ZF.(Since the way of my argument i tried today is exactly the same as the one i tried yesterday, that is, i can't make a countable choice).
I want to prove this if possible, or know whether it is unprovable. Help