Working in $\mathsf{ZF}+\mathsf{DC}$ (that is, we are allowed to use Dependent Choice but not full choice), 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. Also, I am guessing that assuming Dependent Choice should not be so central to the argument, rather it's there mostly just to ensure that one can utilize the usual definitions of Lebesgue measure, measurable, etc. without much trouble.