Yes. It's not the Choice axiom which is problematic, but the infinity itself. So it doesn't mater if ZF or ZFC.
Why do I believe this? It's known for some time now, that you can't have an uniform probability distribution over the set of all naturals. That would be an express road to paradoxes.
The problem is, that even if you have a probability distribution where P(0)=0.5, P(1)=0.25, P(2)=0.125 and so on ... you can then invite a super-task of swapping two random naturals (using this distribution) at the time 0. Then the next swapping at 0.5. Then the next swapping at 0.75 ... and so on.
The question is, what is the probability that 0 will remain in its place? It can't be more than 0, after the completion of the super-task after just a second. On the other hand, for every other number, that probability of being on the leftmost position is also zero.
We apparently can construct an uniform distribution over the naturals. Which is bad.
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What can one expect after this super-task is done to see?
Nothing?
It has been noticed, but never resolved properly. A consensus among top mathematicians, that everything is/must be okay prevails.
One dissident.
https://www.youtube.com/watch?t=27&v=4DNlEq0ZrTo
Phrasing it as a "super-task" relies on intuitions that are not easily formalized in either PA or ZFC. Think instead in terms of a limit, where your nth distribution and let n go to infinity. This avoids the intuitive issues. Then just ask what mean by the limit. You are taking what amounts to a pointwise limit. At this point, what matters then is that it does not follow that a pointwise limit of probability distributions is itself a probability distribution.
If you prefer a different example that doesn't obfuscate as much what is going on we can do it just as well with the reals. Consider the situation where the nth distribution is uniform on the interval from n to n+1. And look at the limit of that (or if you insist move back to having it speed up over time to make it a supertask). Visually what is happening each step is a little 1 by 1 square moving one to the right. Now note that the limit of these distributions is zero everywhere, and not in the nice sense of zero at any specific point but integrates to a finite quantity, but genuinely zero.
This is essentially the same situation, so nothing in your situation has to do with specific aspects of countable sets.