It varies.
If you can prove that there exists a unique $S_i$ then there is no need to use the axiom of choice.
On the other hand, think of $\psi(x) = \exists y(y\in x)$, then every non-empty set has this property, so one can easily construct an example where the axiom of choice is essential.
In the example, since $Y$ is a fixed parameter it is simple to define $G_i=V_i\setminus Y$. You can also take $G_i=V_i$ for all $i$.
For example, if $\cal A=\{A_i\mid i\in I\}$ is a family of non-empty and pairwise disjoint sets then $\varnothing$ can be in at most one element of the family. Without loss of generality, it is in none of the members. If $\cal A$ does not have a choice function then the sentence $\psi(x)=\exists y(y\in x)$ has at least one element in every $A_i$ which satisfy it, in fact $\{a\in A_i\mid \psi(a)\}=A_i$, so we cannot use this to choose from this family.