I have written a formal proof of the theorem:
$$\forall U \exists r(\forall a(a\in r \leftrightarrow (a\in U \wedge a\notin a)) \wedge r\notin U \wedge r\notin r)$$
See: http://www.dcproof.com/SeparationAxiom.htm
(Somewhat non-standard notation: & = $\wedge$, | = $\vee$)
It seems you can select a subset much like the so-called Russell Set from any set $U$ without obtaining a contradiction. How can I be certain?
Edit: Apparently you can't be certain. See my own answer below.