Why this works for proving ZF axioms system is free of Russell's paradox?

This doesn't show that ZF is free from Russel paradox, or consistency of ZF (it's imaginable, though extremely unlikely, that ZF is in fact inconsistent).

This shows that ZF proves that no universal set exists. Reductio ad absurdum works with all theories, even inconsistent: if by adding hypothesis $H$ to theory $T$ we can get inconsistency, then $T$ can prove $\neg H$. Of course it's not interesting if $T$ is itself inconsistent (because then it can prove anything), but it's still correct.

Russell's Paradox arose from an inconsistency in an early attempt to axiomatize set theory by G. Frege. In his system, it was possible to both prove and disprove:

$~~~~\exists x: \forall y: [y\in x \iff y\notin y]$

In ZFC theory, it seems to be impossible to derive this result. You can, of course, derive its negation in ZFC theory.