Sets $A_1,A_2,A_3,...$ with $\dots\in A_3\in A_2\in A_1$ In $\textsf{ZFC}$, is there a sequence of sets $A_1,A_2,A_3, \dots$ such that
$$\cdots \in A_3\in A_2\in A_1?$$
 A: Let me be ornery - the statement is not quite true. There are models of ZFC (assuming it's consistent) in which there are such sets. This is due to the Compactness Theorem (https://en.wikipedia.org/wiki/Compactness_theorem).
What the axiom of foundation (or regularity) rules out is having the whole sequence of sets $\{A_0, A_1, A_2, . . . \}$ exist at once - that is, the model $M$ can't "see" a descending sequence of sets. But it might have a "hidden" descending sequence of sets.
This is discussed here: Can a model of set theory think it is well-founded and in fact not be?
This is exactly analogous to the idea of nonstandard models of arithmetic, where there are infinite descending sequences of "natural numbers" but those sequences can't be defined inside the model, so the induction axioms still hold. See https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic.

EDIT: let me add, for clarity: 
There are two ways to phrase the question you ask.
The first is to ask: "Is there a model $M$ of ZFC, with elements $A_0, A_1, A_2, . . .$, such that $. . . A_2\in A_1\in A_0$?" The answer to this question is "yes," by compactness.
The second is to ask: "Is the sentence 'there is a descending $\in$-sequence' consistent with ZFC?" (More formally: "Is the sentence 'there is a function $f$ with domain $\omega$ such that $f(n+1)\in f(n)$ for every $n\in\omega$' consistent with ZFC?") The answer to this question is "no," by the axiom of foundation/regularity.
This is not an issue you want to focus on too much at first - you should begin by thinking of foundation/regularity as saying "$\in$ is well-founded." In particular, if you read this and find it confusing (which is totally reasonable!), put it aside for later; it's more important that you develop an intuition for what the "nice" (:P) models of set theory look like. But as you think more about set theory, this becomes a really interesting and important point. 
A: Dunno why the down votes, seems like a reasonable question. The answer is no; this is ruled out by the Axiom of Foundation (or Regularity).
A: Let me elaborate a little on David's answer.
Consider the axiom of regularity, which may be stated as:
$$
\forall x (x \neq \emptyset \rightarrow \exists y \in x: x \cap y = \emptyset)
$$
Assume that there is such a sequence $A_1 \ni A_2 \ni \ldots$ and consider the set $\mathcal A = \{A_1, A_2, \ldots \}$. Now try show that this leads to a contradiction.
If you need additional help, just let me now. If you manage to prove it, I encourage you to show your results.
