Separation and Replacement build subsets of sets already in hand, which is exactly what blocks Russell's construction

An unrestricted comprehension principle would assert, for every formula $\varphi(x)$ of the language (The first-order language of set theory: $\in$, $=$, formulas with parameters, and class abbreviations), that $\exists y\,\forall x\,(x \in y \leftrightarrow \varphi(x))$. Taking $\varphi(x) := x \notin x$ turns that assertion into precisely the set There is no $R$ with $x \in R \leftrightarrow x \notin x$ for every $x$ shows cannot exist, so the principle is refutable in plain first-order logic; no appeal to any axiom below is needed to reject it.

The Axiom Schema of Separation: for each formula $\varphi$, $\forall \bar p\,\forall x\,\exists y\,\forall z\,(z \in y \leftrightarrow (z \in x \wedge \varphi(z,\bar p)))$ asserts something weaker in a specific way: it does not produce a set from a formula alone, but only from a formula together with a set $x$ already in hand, and the set it produces is a subset of that $x$. Running Russell's argument against it therefore yields no contradiction but a theorem. Given any set $x$, the separated set $r := \{\, z \in x : z \notin z \,\}$ exists; asking whether $r \in r$ shows that $r \in x$ is impossible, since $r \in x$ would give $r \in r \leftrightarrow r \notin r$. So every set $x$ has a subset that is not one of its elements, and no set contains every set, which is There is no set $U$ with $y \in U$ for every set $y$.

The same restriction is what makes The Axiom Schema of Replacement: for each formula $\varphi$, if $\varphi$ defines a class function on $A$ then its image on $A$ is a set safe. It does not assert that an arbitrary class is a set either: its hypothesis is that a formula behaves like a function on a set $A$ already in hand, and its conclusion is about the image of that particular set. Both schemas therefore build only from material already given, and neither can be turned on the universe at large.

What is given up is small and is worth naming exactly: from these axioms alone nothing whatever can be constructed, which is why The Axiom of Infinity: there is a set containing a set with no elements and closed under $y \mapsto y \cup \{y\}$ is assumed outright rather than derived, and why There is exactly one set with no elements, written $\varnothing$ needs the logical fact that the domain of discourse is nonempty before Separation has anything to act on.