Proving that classes aren't sets I've always had trouble proving that certain classes, (like the universe, $\mathcal W$, or the set of all sets, which I'll call $S$ henceforth). It recently occurred to me that the separation axiom (namely, given a set $A$ and a definite unary condition $P(x)$, $B=\{x \in A \mid P(x)\}$ is also a set) could possibly be used in a proof. For example, 
Assume (towards a contradiction) $S$ is a set, defined as $x \in S \Leftrightarrow \mathrm {Set}(x)$. Applying the unary definite condition $P(x) \Leftrightarrow x \notin x$ to the set yields $T=\{x\in S \mid P(x)\}$, which, by the separation axiom, is a set, but by Russell's paradox, isn't a set.
Is that sufficient to prove that $S$ is not a set?
 A: This does work - for some cases.
For example, you mention specifically the case of showing that the class $S$ of all sets is not a set. Here, this argument does work. However, it will break down if we try to use it to show that e.g. the class of all ordinals is not a set, since the "set" of all ordinals which are not elements of themselves may not contain itself without yielding contradiction - by virtue of not being an ordinal!

Specifically, let's tease out all the details in the proof that the universe, $S$, is not a set:


*

*If $S$ were a set, then $R:=\{x\in S: x\not\in S\}$ would also be a set, by Separation.

*We would then have $R\not\in R$, since if $R\in R$ we must have $R\not\in R$.

*However, $R\in S$ since $S$ is the set of all sets. 

*So $R\in S$ and $R\not\in R$, so by definition $R\in R$; contradiction.
It's that third bullet point that can break down in general, and hence fail to get us a contradiction. If we take $T$ to be the class of all ordinals, and assume for contradiction that $T$ is a set, we can't repeat this argument because it breaks down right here.
EDIT: With some work, we can make this work for the class of ordinals.
Suppose $T$ is a set, which is the set of all ordinals (towards contradiction obviously). By Foundation, no ordinal (or indeed set) is an element of itself, so the set $T':=\{x\in T: x\not\in T\}$ is all of $T$. But then since $T'$ is a transitive set of ordinals, $T'$ is an ordinal (by definition of "ordinal"), so $T'\in T'$, contradiction.
This is of course very silly, but it does technically work. :P
A: EDIT: I believe I misread your post in my original answer (below), but I'm leaving this here as a "comment that's too long to be a comment" because it might be of interest to you or other readers.

Original answer.
That is not a sufficient proof. Russell's paradox relies fundamentally on the fact that, in the specification $\{ x \mid x \not \in x \}$, the variable $x$ is not restricted to a set, but varies over the whole universe of discourse.
Given a set $S$, if you try to mimic Russell's contradiction for the set $T = \{ x \in S \mid x \not \in x \}$, then there are two options:


*

*$T \in T$. In this case, we have $T \in S$ and $T \not \in T$, which is a contradiction.

*$T \not \in T$. In this case, either $T \not \in S$ or $T \in T$. We don't get a contradiction here, because this doesn't force $T \in T$ (which would be contradictory): it simply must be the case that $T \not \in S$!


With Russell's paradox, the option in Case 2 above that $T \not \in S$ doesn't arise, because we're quantifying over the whole universe of discourse.
A: There is no predicate Set$(x)$ in the language of set theory. In many cases we refute $ \exists S\;\forall x\; (P(x)\implies x\in S)$ by showing that it implies $
\exists T\; \forall y\in ON \;\exists z\in T\; (z\in ON \land z>y),$ as this implies $\exists T\;(\cup T\supset ON),$ which in turn implies $\exists U \;\forall w\;(w\in U\iff w\in ON).$  Note that " $ w\in ON $ " is just a convenient abbreviation. 
It is not at all silly to prove that ON is not a set, considering it is a result which is used so much . Even in ZF minus the axiom of foundation the def'n of an ordinal implies that an ordinal cannot belong to itself and hence that $\neg \exists U\;\forall w\;(w\in U\iff w\in ON).$
