For the purposes of this discussion, let a Vitali Set be any subset $V\subseteq{}[0,1)$ such that for $V\_q:=\{x+q\;|\;x<1-q,\;x\in{}V\}\cup\{x+q-1\;|\;x\geq{}1-q,\;x\in{}V\}$ there is a countable subset $I\subset[0,1)$ such that


1. $[0,1)=\bigcup{}\_{q\in{}I}V\_q$
2. For $r,q\in{}I$ distinct, $V\_r\cap{}V\_q=\emptyset$


Can such a $V$ be constructed without AC?



Can a Vitali Set be constructed without AC?
No Vitali set in your sense can be measurable. I am assuming this is the reason for defining a Vitali set in this way.
But Solovay has shown (assuming the consistency of a certain large cardinal, namely an inaccessible) that there is a model of ZF in which all sets of reals are Lebesgue measurable.
In particular, there is no Vitali set in Solovay's model. Hence you need some fragment of AC to construct one.


