Can you construct a Vitali set under ZF? A Vitali set $V$ is a complete set of representatives of $\mathbb{R}/\mathbb{Q}$ in the interval $[0,1]$. Translating $V$ by all rationals in $[-1,1]$ leads to a proof that it is not measurable.
Now, there are models of ZF in which every subset of $\mathbb{R}$ is measurable, so the answer to my question ought to be no.
My confusion is, why does the following approach fail?
Under the assumption that L=V (the constructible universe = the Von Neumann universe), the axiom of choice holds by using a well-ordering of L.
This leads me to suspect that even without L=V, there should be choice functions at least for families of constructible sets, though I might be misunderstanding something.
$\mathbb{R}/\mathbb{Q}$ seems like a constructible set, so why can we not construct a Vitali set in ZF?
 A: It is indeed the case that, if $X$ is a constructible instance of choice (I'm being a bit vague about this, but e.g. maybe take this to mean that $X=(A,E)$ where $A\in L$ is a set and $E\in L$ is an equivalence relation on $A$), then $X$ has a choice function in $V$ even if $V\not=L$. This is because being a choice function/transversal/etc. is absolute between $V$ and $L$, so that anything which $L$ thinks is a witness to choice not failing at $X$ actually is a witness to choice not failing at $X$.
However, this does not let us build a transversal for $\mathbb{R}/\mathbb{Q}$ - rather, it only gives us a transversal for $\color{red}{\mathbb{R}^L}/\mathbb{Q}$, the set of reals in $L$ modulo the rational difference relation (note that $\mathbb{Q}$ and the rational difference relation are each absolute between $V$ and $L$). So indeed we have the following, assuming merely $\mathsf{ZF}$ in $V$: if $\mathbb{R}=\mathbb{R}^L$ then there is a Vitali set. However, this still leaves open the possibility that $V$ fails to have any true Vitali sets by virtue of having (to start with) lots of reals which $L$ doesn't "see."
