The following question arose from a discussion about the definability of bases of $\mathbb{R}$ as a $\mathbb{Q}$vector space. (ZF without AC) something we can note is that the existence of a (definable) wellordering of $\mathbb{R}$ is easily seen to be equivalent to that of a (definable) wellordered basis of $\mathbb{R}$ as a $\mathbb{Q}$vector space. After these remarks, the following question seems natural : Is it consistent with ZF that there be a basis of $\mathbb{R}$ over $\mathbb{Q}$ that cannot be wellordered ?
The problem is generally open. However, recently Liuzhen Wu, Liang Yu, Ralf Schindler and Mariam Beriashvili posted a preprint in which they prove the consistency of the existence of a Hamel basis and $\Bbb R$ cannot be wellordered. Specifically, they show there is such a basis in Cohen's first model.
This can be found on Ralf's homepage.

$\begingroup$ But then what do you mean "generally open" ? Is it simply that they didn't publish yet ? $\endgroup$ – Maxime Ramzi Dec 28 '16 at 10:31

1$\begingroup$ It means that this hasn't been vetted through the usual channels. I read the proof myself, and it looks okay (although I am certain it can be somewhat simplified, at least conceptually, if one takes the approach of symmetric extensions... but then again, I am fluent in the language of symmetric extensions, and less in the language of relative definability.) $\endgroup$ – Asaf Karagila Dec 28 '16 at 11:31
There are actually two papers: "ZF + there is a Hamel basis" doesn't give that every infinite set of reals has a ctble. subset, see https://ivv5hpp.unimuenster.de/u/rds/hamel_basis.pdf , and "ZF + DC + there is a Hamel basis" doesn't give that there is a w.o. of R, see https://ivv5hpp.unimuenster.de/u/rds/hamel_basis_2.pdf