For any set $A$, there is some $x$ not in $A$. I am working through a set theory text and am having trouble giving a formal proof of the above statement. So far I have:
Suppose the contrary. Then there exists a set $A$ such that $x\in A$ for all $x$. Thus $A$ is the set of all sets, which gives rise to Russell's paradox.
Is there a more concise or direct proof of this fact using the axioms of $\sf ZFC$? I am concerned about asserting that $A$ is the "set of all sets."  

A more direct (but very similar) proof can be concieved: even though Russell's paradox is often thought as a limitative result, the argument it uses is precisely a proof of your result:
Let $x$ be the subset $\{a \in A  \ | \ a \notin a\}$ of $A$. $x \in A$ is absurd, hence $x \notin A$.
Note that assuming fondation, $x$ is equal to $A$.
