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 :
$\lnot \exists y \forall x((x \in y) \iff (x \notin x))$.
Jech makes two claims here:
The idea behind comprehension is not paradoxical.
The idea of a universal set is.
The key word doing all the heavy lifting in idea 1 is "idea:" for Jech, the full comprehension scheme and the full separation scheme are each implementations of the comprehension idea, and the fact that the former (which is more obvious) is inconsistent doesn't mean that the underlying idea itself is inconsistent. Note that there's an implicit claim here that separation does not represent a new idea but rather a more careful formulation of the same idea behind comprehension. I tentatively agree with Jech here.
Claim 2, to be honest, I have little sympathy for. Certainly there are consistent set theories which admit a universal set (e.g. $\mathsf{NFU}$ - while the consistency of $\mathsf{NF}$ relative to $\mathsf{ZF}$ is (sorta-kinda-)open, it's easy to show that $\mathsf{NFU}$ is consistent relative to even much less than $\mathsf{ZF}$).
One possible response to my objection to claim 2 is that theories like $\mathsf{NFU}$ while technically having a universal set aren't faithful to the idea behind a universal set - that part of sethood is separation. If we grant this, then it is indeed the case that "the idea of a universal set" is contradictory; however, this goes against the putative bundling of separation and comprehension two paragraphs prior.
Ultimately, I disagree with Jech here: I think there are two genuinely different ideas, each of which is consistent on its own but which are mutually inconsistent.
Best Answer
The magnificent beauty of the axioms of $\sf ZF$ is that they allow us, with only $\in$ to express so much.
When we say an arbitrary property, we mean one that can be expressed in the language of set theory. Otherwise it will be impossible to write the axiom relevant to that property in the language of set theory, which is the language of $\sf ZF$.
You seem to forget that there are quantifiers to be used in the formulas. Not everything is boolean combinations of atomic formulas and their negations. No, we make heavy use of quantification here. For example $\subseteq$ can be defined as $x\subseteq y\iff\forall z(z\in x\rightarrow z\in y)$, and we can define when a set is transitive, $\forall y(y\in x\rightarrow\forall u(u\in y\rightarrow u\in x))$, or in shorter form, $\forall y(y\in x\rightarrow y\subseteq x)$.
We can define $x\cup y$, and $x\cap y$, and more and more. Some of the formulas require us to rely on the axioms in order to prove their correctness, but that's fine. We are allowed to do that. But we can sit and write formulas which quickly become more and more complicated and those express a lot. A lot more than just $x\in y$ or $x=y$.