The Stacks project

Proof. Note that if $Y'$ is affine, then $X'$ is affine as $\pi$ is finite. Choose an affine open neighbourhood $U' \subset S'$ of $s'$. Choose an affine open neighbourhood $V' \subset h^{-1}(U')$ of $y'$. Let $W' = h(V')$. This is an open neighbourhood of $s'$ in $S'$, see Morphisms, Lemma 29.34.10, contained in $U'$. Choose an affine open neighbourhood $U'' \subset W'$ of $s'$. Then $h^{-1}(U'') \cap V'$ is affine because it is equal to $U'' \times _{U'} V'$. By construction $h^{-1}(U'') \cap V' \to U''$ is a surjective smooth morphism whose fibres are (nonempty) open subschemes of geometrically integral fibres of $Y' \to S'$, and hence geometrically integral. Thus we may replace $S'$ by $U''$ and $Y'$ by $h^{-1}(U'') \cap V'$. $\square$