This is not an open problem anymore.
In 1980 Moti Gitik published a solution. Starting with a proper class of strongly compact cardinals, one can construct a model of $\sf ZF$ in which every limit ordinal has countable cofinality. Therefore there are no uncountable regular $\aleph$'s.
M. Gitik, All uncountable cardinals can be singular, Israel J. Math. 35 (1980), no. 1-2, 61--88.
The informed guess comes from understanding the Feferman-Levy model in which $\omega_1$ is a countable union of countable ordinals (i.e. has a countable cofinality).
While one can show without using the axiom of choice that for an infinite ordinal $\alpha$ it holds that $|\alpha|=|\alpha\times\alpha|$, to use that to show that $\omega_1$ is regular, one has to take a countable list of countable ordinals and prove their union is countable. We do that by choosing enumerations for these ordinals, and then the union is an enumerated union, which is indeed countable.
But as the Feferman-Levy model showed, one has to use the axiom of choice to choose these enumerations. The result translates to larger cardinals as well.
And so Gitik's work shows that indeed assuming large cardinals this is impossible to boot. One might point out that large cardinals are necessary. It was [recently] shown that if $\kappa$ and $\kappa^+$ are singular then there is an inner model with a Woodin cardinal. So to have all uncountable $\aleph$ singular, one would have to use at least a proper class of Woodin cardinals, and we expect this to be even more than just that.
Ralf-Dieter Schindler, Successive weakly compact or singular cardinals, J. Symbolic Logic 64 (1999), no. 1, 139--146.