Lebesgue Measurable Sets & Axiom of Determinacy

While reading some logic theory I bumped against the theorem which states that every set of reals is Lebesgue measurable, assuming the axiom of determinacy. To prove this theorem it apparently suffices to prove the following lemma.

Lemma. Assuming AD, let $S$ be a set of reals such that every measurable $Z \subset S$ is null. Then $S$ is null.

The reasoning why it suffices to prove this lemma goes as follows. Let $X$ be a set of reals. Let $A \supset X$ be a measurable set with the property that every measurable $Z \subset A-X$ is null. Then $A-X$ is null and hence $X$ is measurable.

Can anyone explain this reasoning to me since I'm a little rusty with my measure theory.

At first, why does such an $A$ exist? And secondly, why does $A-X$ is null imply that $X$ is Lebesgue measurable?

Question 1,

suppose $X$ is not measurable.

let $a = m^*(X)$, then there exist a Borel set $A\supseteq X$ , $m(A)=m^*(A)=m^*(X)=a$ (by definition of outer measure).

if $K\subseteq A-X$ is $measurable$ and $m(K)>0$, then $A-K$ covers $X$.

so, $a=m^*(X)\le m^*(A-K) = m(A-K) = m(A)-m(K)<m(A)=a$ , contradiction.

Question 2,

if $m(A-X)=0$. $A-X$ is null,so measurable, then $X= A - (A-X)$, measurable.