TITLE: Prove that the class of well-founded sets is a proper class
QUESTION [5 upvotes]: Does anyone have an "elementary" proof of the following claim:

If $A$ is a class such that
  $$(*)\qquad\forall x(x\subseteq A\to x\in A),$$ then $A$ is a proper class, i.e. $\forall y\ y\ne A$.

The reason that the title refers to well-founded sets is that if we define the standard cumulative hierarchy $V_\alpha=\bigcup_{\delta<\alpha}{\cal P}(V_\delta)$ with ${\rm WF}=\bigcup_{\alpha\in{\sf On}}V_\alpha$, then it can be proven that $A\supseteq{\rm WF}$ if $A$ satisfies $(*)$. Now I know a proof of the claim using this and ${\rm WF}\supseteq{\sf On}$, together with the Burali-Forti paradox and the subset axiom, but that seems a bit contrived. I feel like there should be a simple proof by contradiction in the vein of Russel's paradox or the Burali-Forti paradox. (Don't assume regularity; it is not necessary, but the proof is trivial assuming it, because any set satisfying $(*)$ is an element of itself.)

REPLY [2 votes]: Theorem. For every set $a$ there is a set $b\subseteq a$ such that $b\notin a$.
Proof. Let $b=\{x\in a:x\notin x\}$.