Statements Equivalent to Axiom of Foundation Yes, this is homework. We're asked to show that the following 3 statements for the Axiom of Foundation are equivalent:
$(1)\quad V = \bigcup_{\alpha} V_{\alpha}$
 $(2)\quad \in \text{ is well-founded.}$
 $(3)\quad \text{If } x \text{ is a non-empty set, then there is an } a \in x \text{ such that } a \cap x = \varnothing$
I was able to rewrite $(3)$ to show that it claims that there's an $\in$-minimal element $a$ in every non-empty set $x$, implying statement $(2)$. I was also able to show that $(2)$ implies $(3)$ by contradiction after invoking an infinite descending $\in$-chain.
I also showed that $(1)$ implies $(3)$, but what I'm having a hard time doing is showing that $(3)$ implies $(1)$ or $(2)$ implies $(1)$.
Any hints?

First, you want to prove that if $m$ is a set and $m\subset\bigcup_\alpha V_\alpha$, then $m\in \bigcup_\alpha V_\alpha$.
Once you have that, you might be tempted to make the following argument.  Let $S=V\setminus\bigcup_{\alpha} V_\alpha$.  If $S$ is nonempty, then (2) says it has an $\in$-minimal element $m$.  But then every element of $m$ is in $\bigcup_\alpha V_\alpha$, so by the previous paragraph, $m\in \bigcup_\alpha V_\alpha$, which is a contradiction.
Unfortunately, this argument doesn't quite work, because $S$ might not be a set (a priori, it is only a class).  To get around this, use the following trick.  Suppose $x\in V\setminus \bigcup_\alpha V_\alpha$, and let $T$ be the transitive closure of $\{x\}$.  Let $S=T\cap (V\setminus \bigcup_\alpha V_\alpha)$ (this is a set by Separation since $T$ is a set).  Now apply the argument above to this $S$.
