The Stacks project

Proof. Writing $S_ i$ as a finite union of open affine subschemes reduces the question to the case that $S_ i$ is affine and $E$ is constructible, see Lemma 32.2.2 and Properties, Lemma 28.2.1. In this case the complement $S_ i \setminus E$ is constructible too. Hence there exists an affine scheme $T$ and a morphism $T \to S_ i$ whose image is $S_ i \setminus E$, see Algebra, Lemma 10.29.4. By Lemma 32.4.9 we see that $T \times _{S_ i} S_{i'}$ is empty for all sufficiently large $i'$, and hence $f_{i'i}(S_{i'}) \subset E$ for all sufficiently large $i'$. $\square$