It's well known that the full axiom of choice is not needed to prove the existence of non-measurable subsets of $\mathbb{R}$. In particular, the Boolean prime ideal theorem ($\mathsf{BPI}$) is sufficient, since non-principle ultrafilters on $\omega$, coded as a subset of $2^\omega$, are not Lebesgue measurable. They also necessarily fail to be Baire measurable.
That said, they can satisfy a weaker regularity condition. I've convinced myself that under $\mathsf{ZFC}+\mathsf{CH}$ if $\mathcal{F}_0 \subset 2^\omega$ is a proper filter generated by an $F_\sigma$ subset of $2^\omega$, then $\mathcal{F}_0$ can be extended to an ultrafilter $\mathcal{F}$ with the property that for every Borel set $B \subseteq 2^\omega$, both $B \cap \mathcal{F}$ and $B \setminus \mathcal{F}$ have the perfect set property (i.e., are either finite, countable, or have a non-empty perfect subset). I'm not sure about filters generated by Borel sets in general, and I do not know what $\mathsf{ZFC}$ entails regarding the question of whether every ultrafilter on $\omega$ has this property.
This has lead me to the following question.
Question. Is $\mathsf{ZF}+\mathsf{BPI}+\text{“Every set of reals has the perfect set property."}$ consistent?
Looking through Consequences of the Axiom of Choice by Howard and Rubin, it seems like this was open 24 years ago at the very least.
