It is easy to prove the following in Z+CC (Zermelo plus countable choice):

Every uncountable closed set of reals is in bijection with the reals.

I was informed by Asaf that it can be proven in ZF (no choice at all), but that proof appears to use replacement. I hence asked whether it could be proven in just Z, but till today there has been no answer. And whether the answer is yes or no, it would be very interesting. If yes, then the proof is likely to be far from obvious, maybe even not previously known. If no, then we have a theorem that needs either choice or replacement over Z, despite those two principles seeming to be completely unrelated.