If we assume the existence of a set $R$ such that $$R=\{x: x\notin x\}$$then we can obtain the contradiction $R\in R $ and $R\notin R$. So, $R$ cannot exist.
To avoid Russell's Paradox then, all we need to do is not assume that, for every unary predicate $P$, there exists a set $S$ such that $$S=\{x:P(x)\}$$
If, however, $A$ is assumed or proven to be a set, then we can assume without fear of contradiction that there exists a subset $S\subset A$ such that $$S=\{x\in A: P(x)\}$$Or equivalently$$S=\{ x: x\in A \text{ and } P(x)\}$$
You can think of $P(x)$ as the criterion for selecting elements from the set $A$ for the subset $S$. The only restriction is that the variable $S$ may not occur in the selection criterion. This is the Axiom of Specification (Selection).
If, for example, we have set $A$, then we can assume that there exists a subset $S\subset A$ such that $$S=\{x\in A: x\notin x\}$$ Or equivalently $$S=\{x:x\in A \text{ and } x\notin x\}$$ Then we would not obtain a contradiction, but we would have $S\notin A$. (Proof left as an exercise.)