See Russell's Paradox :
Zermelo replaces NC [Naïve Comprehension principle] with the following axiom schema of Separation (or Aussonderungsaxiom):
$$∀A ∃B ∀x (x \in B \iff (x \in A \land \varphi)).$$
Again, to avoid circularity, $B$ cannot be free in $\varphi$. This demands that in order to gain entry into $B$, $x$ must be a member of an existing set $A$. As one might imagine, this requires a host of additional set-existence axioms, none of which would be required if NC had held up.
How does Separation avoid Russell's paradox? One might think at first that it doesn't. After all, if we let $A$ be $V$ – the whole universe of sets – and $\varphi$ be $x ∉ x$, a contradiction again appears to arise. But in this case, all the contradiction shows is that $V$ is not a set. All the contradiction shows is that “$V$” is an empty name (i.e., that it has no reference, that $V$ does not exist), since the ontology of Zermelo's system consists solely of sets.
Consider a set $z$, we can still form the set $R = \{ x \in z : x \notin x \}$;
This only implies that :
if $R \in z$, then $R \in R \ \text { iff } \ R \notin R$, which means (reductio ad absurdum) that $R \notin z$.
Thus :
there is no universal set : $\forall z \exists R(R \notin z)$.
It can be worth to note that the "non-existence" of the Russell's set $R$ can be proved by logic alone :
1) $\exists y \forall x(A(x,y) \iff \lnot A(x,x))$ --- assumed [a]
2) $\forall x(A(x,c) \iff \lnot A(x,x))$ --- with $c$ a new constant
3) $A(c,c) \iff \lnot A(c,c)$ --- by instantiation.
The last line gives us a contradiction, because : $\vdash A(c,c) \iff A(c,c)$; thus, we conclude with :
$\vdash \lnot \exists y \forall x(A(x,y) \iff \lnot A(x,x))$ --- (*)
discharging the assumption [a].
Now, if we apply (*) to the language of set theory with the binary predicate $\in$ in place of $A$, we get :
$\lnot \exists y \forall x((x \in y) \iff (x \notin x))$.
Not quite; you must still consider why the original contradiction $R \in R \iff R \not \in R$ no longer applies.
That is because once you limit to $B$, you can no longer conclude $R \in R \iff R \not \in R$ anymore; you can only conclude that under the additional assumption that $R \in B$. So the conclusion is simply that $R \not \in B$, and there is no contradiction.
Note that this holds even if you remove the axiom of regularity.
Best Answer
The guarantee that such a set can't exist is already given by the argument of Russell's paradox: its existence leads to a contradiction therefore it can't exist.
The problem with unrestricted comprehension was that it guaranteed the set does exist, which causes a problem because of the conflicting guarantees.