If $ZF$ is consistent, then since there is a model of $ZF+\mathrm{cf}(\omega_{1})=\omega_{0}$, $ZF\not\vdash\mathrm{cf}(\omega_{1})=\omega_{1}$, and since $ZFC$ is consistent, $ZF\not\vdash\mathrm{cf}(\omega_{1})\not=\omega_{1}$.
Can it be proved in $ZF$ that there is an uncountable regular cardinal unless $ZF$ is not consistent?
Perhaps surprisingly, it (probably) cannot!
Gitik showed that there is a model of $\mathsf{ZF}$ in which every cardinal is singular, assuming that $\mathsf{ZFC}$ + "There is a proper class of strongly compact cardinals" is consistent. This consistency hypothesis is quite strong. By contrast, the "classical" independence results ($\mathsf{ZF}$ can't prove or disprove $\mathsf{AC}$, $\mathsf{ZFC}$ can't prove or disprove $\mathsf{CH}$, etc.) have no consistency hypothesis beyond the obvious one of the consistency of the "base theory" itself.
This raises the question of whether this consistency hypothesis can be removed - or, put another way, what else must be consistent if $\mathsf{ZF}$ + "Every uncountable cardinal is singular" is consistent. My recollection is that Gitik's hypothesis is now known to be far from optimal, but I can't find a citation for that at the moment - I'll add it if/when I track it down. (Or Asaf Karagila will post it about $30$ seconds after I finish typing this.)
EDIT: these slides of Seveliev indicate on page $16$ that the answer is a proper class of Woodin cardinals, but no specific source is given. So I'm still looking. (That said, those slides do seem quite relevant and interesting on their own so they're very much worth mentioning.)
