I'm trying to argue that a Lebesgue non-measurable subset $X$ of $\mathbb{R}^n$ has a non-measurable subset $Y$ such that any measurable subset of $Y$ has measure zero.
Let $A_0$ be a measurable subset of $X$. If all such $A_0$ has measure zero, then let $Y = X$ and we are done. Otherwise, choose $A_0$ which has positive measure. By Vitali construction, there exists a non-measurable subset $A_1 \subset A_0$. Again, if $A_1$ doesn't have a measurable subset of positive measure, then set $Y = A_1$. Otherwise, we repeat what we did as above.
How does one rigorously conclude that the process will have to eventually terminate?
Best Answer
The following turns my comments above into an answer (modulo exercises :P).
Throughout, I'll assume that $X$ is a non-measurable set which has a positive-measure measurable subset.
The process in the OP is too loosely-defined to be guaranteed to terminate - maybe we just happen to keep grabbing "inefficient" subsets of our starting set. Instead, it's better to try to carve out a "maximal(ish) measurable subset" (and think about what's left over).
Note my careful use of the suffix "ish" in the above. There is not going to be a literally maximal measurable subset of $X$: if $A\subseteq X$ is measurable and $a\in X\setminus A$, then $A\cup\{a\}$ is also a measurable subset of $X$. Similarly, we can't do something like taking the union of all the positive-measure subsets of $X$: that will just yield $X$ itself (again thinking about how singletons don't affect measurability), which isn't measurable. This second obstacle brings up a related issue, namely that measurability is fragile: if I take a union of a bunch of measurable sets, the result may not be measurable. Only countable unions are safe: the union of countably many measurable sets is always guaranteed to be measurable.
So we want to find a measurable subset of $X$ which in some sense is as big as possible, but $(i)$ that sense has to be somewhat nuanced and $(ii)$ we have to only take a small number (= countably many) of unions of measurable pieces. What can we do?
Well, besides literal maximality, we do have at hand a way of measuring when something is as big as possible: measure! Say that $A\subseteq X$ is a quasimaximal-measurable subset of $X$ (not an actual term) iff $A$ is measurable and, for every measurable $B$ with $A\subseteq B\subseteq X$, we have $m(A)=m(B)$ (or equivalently, if $B$ is any measurable subset of $X$ we have $m(A)\ge m(B)$). So maybe $A$ isn't literally as big as possible, but we can't do any better measure-wise.
Using basic facts about measurability, you should be able to prove the following statements:
If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ is non-null.
If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ does not have a positive-measure measurable subset.
At this point you're basically done ... except that you need to prove that quasimaximal-measurable subsets exist at all! HINT: consider the number $$\alpha=\sup\{m(A): A\subseteq X\mbox{ is measurable}\}$$ (for simplicity let's assume $\alpha<\infty$). Can you see how to build a measurable subset of $X$ with measure exactly $\alpha$ (think about countable unions!), and why such a subset must be quasimaximal-measurable?