Working only on $\mathsf{ZF}$ (without choice, that is), suppose that there exists a non-measurable subset of the unit interval $[0,1]$ (just non-measurable, so we're not allowed to assume that this is a Berstein set, or a Vitali set, or any other "concrete" kind of non-measurable set). How can one prove that there must exist a non-measurable subset $X\subseteq[0,1]$ such that $\mu_*(X)=0$ and $\mu^*(X)=1$?

For context: I am attempting to understand the proof that the Axiom of Determinacy implies every subset of $\mathbb R$ is Lebesgue-measurable. The strategy (as it appears in various sources) seems to be, given an $X\subseteq[0,1]$, to define a game such that a winning strategy for one player implies $\mu_*(X)>0$ while a winning strategy for the other player implies $\mu^*(X)<1$. So a proof like this necessarily uses the fact that the existence of some non-measurable set implies that there exists another with inner measure zero and full outer measure; all the proofs I've read simply mention this as if it was obvious, but I am finding it nontrivial to come up with a proof.