a)Prove that a "set of all sets" does not exist. [Hint: if $V$ is a set of all sets, consider $\{x\in\,V\,|\;x\notin x\}$.
b)Prove that for any set $A$, there is some $x\notin A$.
These exercises are from chapter 1, section 3 of Jech and Hrbacek's "Introduction to Set Theory." For some reason, I'm struggling to axiomatically show a). It seems b) will naturally follow course afterward, simply employing proof by contradiction, essentially making it part a).
a) Let $V$ be a set of all sets. In other words, $\forall x, x\in V$. Since $V$ is, by hypothesis, a set, then by comprehension schema axiom: for the property $\mathbf P(x)$ given by "$x\notin x$", there exists some set, $C$, such that $x\in C$ if and only if $x\in V$ and $\mathbf P(x)$. Now, if $V\in C$, this implies $V\in V$ and $V\notin V$, a clear contradiction. Hence, $V\notin C$. But by the definition of $C$, this, in turn, implies $\lnot \bigl(V\in V$ and $\mathbf P(V)\bigl)$, which means $V\notin V$ or $\lnot \mathbf P(V)$; in other words, $V\notin V$ or $V\in V$. Suppose $V\notin V$. Then, at the same time, by the definition of $V$, $V\in V$; a contradiction. Now suppose $V\in V$...? [can't seem to be able to reach contradiction from here]