Let x be a set. Then, V = {x| x not in x} is the set that causes Russel's Paradox. We can easily define a new set S = {x| p(x) and x in U} where p(x) is some property and U is the universal set. Then S easily fixes the contradiction.
Edit: @mafribe makes the point that there are some set theories that can still have universal sets by culling other features that ZF-style set theories have. I was mostly referring to ZF-style set theory (hence my citation). Indeed, one could even make a ZF-style set theory paraconsistent and still have Universal sets.
[1] http://sistemas.fciencias.unam.mx/~lokylog/images/stories/Al...
Having a universal set is a sufficient condition for Russel's paradox.
[1] http://plato.stanford.edu/entries/quine-nf/
[2] https://www.dpmms.cam.ac.uk/~tf/
[3] T. E. Forster, Set Theory with a Universal Set. http://ukcatalogue.oup.com/product/9780198514770.do
Your [3] link doesn't work by the way, I'm interested in reading Forster!
If V = {x | x not in x}, then if V contains itself, V is not in V (and vice versa) is an obvious contradiction. Your new set S doesn't help in the slightest.
Fixing it is emphatically "not easy" and mathematicians generally rely on the ZFC axiomation (although several other possibilities were proposed).
But now you're talking about non-standard set theory[1] which is fine but you are kind of side-stepping the issue.
If you define S to be S = {x in U: x not in x}, then it simply means S is not in U.