As far as I understand, it is consistent with ZF that every set of real numbers is a Borel set. Furthermore, I know that relatively weak forms of the axiom of choice suffice to prove that $|\mathcal B| = \mathfrak c$. The standard proof that constructs the Borel hierarchy requires the regularty of $\omega_1$ (so, for example, countable choice) to show that the hierarchy collapses at $\omega_1$, and then requires -- as far as I can tell -- something like the well-orderability of $\mathbb R$ to show that $\aleph_1 \times \mathfrak c = \mathfrak c$.
Can we relax this last 'requirement'? In particular, can we show that there are continuum-many Borel sets using only countable choice?