Proof.
Assumptions (1) and (2) imply that S' \to S is a universal homeomorphism (for example because S_{red} = S'_{red} and using Lemma 29.45.6). Hence (3) implies that X \to S is homeomorphism onto a closed subset of S. Then X \to S is affine by Lemma 29.45.4. Let U \subset S be an affine open, say U = \mathop{\mathrm{Spec}}(A). Then S' = \mathop{\mathrm{Spec}}(A/I) by (1) for a locally nilpotent ideal I by (2). As f is affine we see that f^{-1}(U) = \mathop{\mathrm{Spec}}(B). Assumption (4) tells us B is a finite type A-algebra (Lemma 29.15.2) or that I is finitely generated (Lemma 29.21.7). Assumption (3) is that A/I \to B/IB is surjective. From Algebra, Lemma 10.126.9 if A \to B is of finite type or Algebra, Lemma 10.20.1 if I is finitely generated and hence nilpotent we deduce that A \to B is surjective. This means that f is a closed immersion, see Lemma 29.2.1.
\square
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