Let $M$ be a countable set. I can take a finite set $F$ of n-ary operations and construct a minimal set $M'$ for which:
- $M \subset M'$
- For each $f \in F$ and $m_1,m_2…m_n \in M$ where $n$ is the arity of $f$: $f(m_1,m_2…m_n) \in M$
$M'$ is a set which is closed against all operations in $F$. It can be shown that $M'$ is also countable. Each element of $M'$ can be described by a finite string using elements from $M$ and operations from $F$.
Now let $M$ be $\Bbb{Q}^3$ and $F$ be a set of all rotations needed in Banach-Tarski paradox (i.e. four rotations) and translation. Then ${\Bbb{Q}^3}'$ is a countable set. Now I can perform the Banach-Tarski paradox in this space the same way as in $\Bbb{R}^3$ with one exception: I don't need the axiom of choice because I use only countable sets. ${\Bbb{Q}^3}'$ is countable so it can be well ordered. For each subset of ${\Bbb{Q}^3}'$, I can pick the least element of this ordering.
Is there anything I missed?