Set with no infinite decending chains is well founded?

Question

I want to show that if a set $x$ doesn't have an infinite decending chain of sets as $x \ni x_1 \ni x_2 \ni x_3 ...$ then it is well founded, meaning that it belongs to $V=\bigcup_{\alpha \in ON} V(\alpha)$ (the Von-Neumann universe). If I try by contradiction and suppose $x \notin V$ so $x \notin V(\alpha) \forall \alpha \in ON$ but then I don't really know how to proceed.

Answer 1

Asaf's comment suggests that you need the axiom of choice for this. I would take his word for it.

Now, recall (or prove as an exercise) the following fact: a set $y$ is in $V$ if and only if all its elements are in $V$.

Hence, if $x$ is not in $V$, then it must contain an element $x_1$ which is not in $V$. Then $x_1$ must in turn contain an element $x_2$ which is not in $V$. This progression cannot terminate, so (using the axiom of choice) we get a sequence $x \ni x_1 \ni x_2 \ni \dots$ which is an infinite descending chain.