How much choice is needed to prove this statement?

Question

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(C_\alpha, x_\alpha)_{\alpha \in \omega_1}$ s.t. $C_\alpha \subseteq \mathbb{N}^\mathbb{N}$ is closed and $\text{CB}^\alpha(C_\alpha) = \{x_\alpha\}$

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers) with the usual topology and $\text{CB}^\alpha$ is the $\alpha$-th Cantor-Bendixon derivative.

My questions are:

1. How much choice do we need at least to prove this statement? Or more generally what known weak axiom can be assumed (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) to prove the statement?
2. Is it consisent $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R}) + \neg$The above statement?
3. If we assume the consistency of $\mathsf{ZF}+\mathsf{AD}$, is it consistent $\mathsf{ZF}+\mathsf{AD}\ +$ The above statement?

Thanks!

Answer 1

Note that these are all countable sets, so what you'd have here is a sequence of countable sets of reals indexed by $\omega_1$.

This is certainly inconsistent with $\sf AD$, since from such a sequence we can construct a subset of size $\aleph_1$. Therefore this is also consistently false with $\sf AC_\omega(\Bbb R)$, at least assuming an inaccessible, as that's the case in Solovay's model.