Why Russell paradox is not treated as negative proof?

In naive set theory, we claim that it is possible to construct sets by saying which elements it contains and which not (all elements from universum have to satisfy some sort of rule or predicate to be in set). And the definition of Russel's set is completly right in naive set theory, because it says clearly, which elements it contains and which not. Since it's well defined in naive set theory, but leads to a contradiction, we obtain a paradox.

Naive set theory was (for the most part) centred around the notion of Unrestricted Comprehension: Given any property $\mathcal{P}$ there exists a set whose elements are precisely the objects $x$ satisfying property $\mathcal{P}$. A perfectly good mathematical property would be "$x$ is a set and is not an element of itself," and so there must be a set $R$ whose elements are exactly those sets which are not elements of themselves. This is Russell's paradoxical set.

So $R$ proved that Unrestricted Comprehension was inconsistent, and we thus have to disallow the construction of this "set" $R$, but still allow for the construction of interesting sets. We could, of course, simply say that Unrestriction Comprehension is fine, except for this set. However you then can still arrive at the Burali-Forti paradox. So maybe we also disallow the the construction of this problematic set. But there will be yet more problematic sets. At any rate, it would seem to be impossible to prove that by simply disallowing the construction of sets appearing in some list of known "paradoxical properties" that the resulting theory becomes consistent.

Much better would be to describe principles of set construction which would seem to disallow the construction of these paradoxical sets. The most common axiomatization now, ZFC, has the following result:

Fact: If ZFC is consistent, then there is no set of all sets. (This is true even for a very small fragment of ZFC.)

Proof: If there were a set of all sets, call it $V$, then by the Axiom Schema of Separation we can construct the set $W = \{ x \in V : x \notin x \}$. However as $W$ is a set, then $W \in V$, and it follows that $W \in W$ iff $W \notin W$! $\Box$

Unfortunately, due to certain amazing theorems of Kurt Gödel, we cannot prove that ZFC is consistent (without transcending ZFC itself) and so there is no proof that ZFC does in fact avoid all paradoxical sets. But several decades of intense research has not yielded any contradictions, so it is fair to say that most set-theorists believe in the consistency of ZFC.