The Axiom of Regularity prevents quine atoms like $A=\{A\}$ because the intersection between $A$ and $\{A\}$ is non-empty.
However, I can't seem to find any contradiction with $A=\{\{A\}\}$ since the axiom of regularity only reaches one set containment deep.
If this set does violate the axioms, how so? If not, what is the point of the axiom of regularity? Would this be a Quine atom in ZFC?
Thanks in advance!
Assume $\{\{A\}\}\in A$, then consider the following set $$B=\{A,\{A\},\{\{A\}\}\}$$ Is there an object of $B$ disjoint from it? Can you generalize this? That is, assume $A_{n-1}=\{A_n\}$ and $A_0=A$, can you generalize the definition of $B$ to prove $A_n\in A$ cannot be true?
The first use of Regularity by Zermelo only ruled out $A\in A$ and so Russell's paradox. The form we have now is to block infinite descending $\in$-chains $$A_0\ni A_1\ni\cdots\tag{*}$$since it is the consensus that only well-founded sets (sets without infinite descending $\in$-chains like $(*)$) are "all sets". I suggest you take a look at the article Believing the Axioms if you want to know more about this.
Edit Allow me to correct my bad phrasing. Disclaimer: I'm about to quote the article above but I can't say I'm 100% sure that is true. If there are historical errors in what I'm about to say, I'll gladly correct them. Anyway, as the article says
As Vsotvep rightfully points out, ZF without Regularity is already enough to block Russell's paradox. The article, in fact, continues with
So the correct phrasing in my answer should have been "The first use of Regularity by Zermelo (in the form $A\notin A$) aimed at blocking Russell's paradox". I wasn't claiming that Regularity ruled out Russell (although I understand why it might seem that way), but that the first instance of a "Regularity"-like axiom aimed at that.