The Stacks project

Proof. We will denote $T \mapsto T_0$ the base change by $\mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A)$. By a weak version of Chow's lemma (in the form of Cohomology of Spaces, Lemma 69.18.1) there exists a surjective proper morphism $\varphi : X' \to X$ such that $X'$ admits an immersion into $\mathbf{P}^ n_ A$. Set $Y' = \varphi ^{-1}(Y)$. This is an open and closed subscheme of $X'_0$. The lemma holds for $(X', Y')$ by More on Morphisms, Lemma 37.53.9. Let $W' \subset X'$ be the open and closed subscheme proper over $A$ such that $Y' = W'_0$. By Morphisms of Spaces, Lemma 67.40.6 $Q_1 = \varphi (|W'|) \subset |X|$ and $Q_2 = \varphi (|X' \setminus W'|) \subset |X|$ are closed subsets and by Morphisms of Spaces, Lemma 67.40.7 any closed subspace structure on $Q_1$ is proper over $A$. The image of $Q_1 \cap Q_2$ in $\mathop{\mathrm{Spec}}(A)$ is closed. Since $(A, I)$ is henselian, if $Q_1 \cap Q_2$ is nonempty, then we find that $Q_1 \cap Q_2$ has a point lying over $\mathop{\mathrm{Spec}}(A/I)$. This is impossible as $W'_0 = Y' = \varphi ^{-1}(Y)$. We conclude that $Q_1$ is open and closed in $|X|$. Let $W \subset X$ be the corresponding open and closed subspace. Then $W$ is proper over $A$ with $W_0 = Y$. $\square$