The Stacks project

Proof. It is harmless to replace $X$ by an open neighbourhood of $x$ in $X$ (small detail omitted). By Properties, Lemma 28.29.8 we may assume $X$ is affine with $\Gamma (X, \mathcal{O}_ X) \subset \mathcal{O}_{X, x}$. In particular $X$ is integral with a unique generic point $\xi$ whose residue field is the fraction field $K$ of the valuation ring $\mathcal{O}_{X, x}$. Since $x$ is in the closure of $U$ we see that $U$ is not empty, hence $U$ contains $\xi$. Thus by the valuative criterion of properness (Lemma 29.42.1) there is a morphism $t : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to Y$ fitting into a commutative diagram

of morphisms of schemes over $S$. Applying Lemma 29.42.4 with $y = t(x)$ and $\varphi = t^\sharp _ x$ we obtain an open neighbourhood $V \subset X$ of $x$ and a morphism $g : V \to Y$ over $S$ which sends $x$ to $y$ and such that $g^\sharp _ x = t^\sharp _ x$. As $Y \to S$ is separated, the equalizer $E$ of $f|_{U \cap V}$ and $g|_{U \cap V}$ is a closed subscheme of $U \cap V$, see Schemes, Lemma 26.21.5. Since $f$ and $g$ determine the same morphism $\mathop{\mathrm{Spec}}(K) \to Y$ by construction we see that $E$ contains the generic point of the integral scheme $U \cap V$. Hence $E = U \cap V$ and we conclude that $f$ and $g$ glue to a morphism $U' = U \cup V \to Y$ as desired. $\square$