**Proof.**
Assume $X$ and $Y$ are birational. Let $f : U \to Y$ and $g : V \to X$ define inverse dominant rational maps from $X$ to $Y$ and from $Y$ to $X$. We may assume $V$ affine. We may replace $U$ by an affine open of $f^{-1}(V)$. As $g \circ f$ is the identity as a dominant rational map, we see that the composition $U \to V \to X$ is the identity on a dense open of $U$. Thus after replacing $U$ by a smaller affine open we may assume that $U \to V \to X$ is the inclusion of $U$ into $X$. It follows that $U \to V$ is an immersion (apply Schemes, Lemma 26.21.11 to $U \to g^{-1}(U) \to U$). However, switching the roles of $U$ and $V$ and redoing the argument above, we see that there exists a nonempty affine open $V' \subset V$ such that the inclusion factors as $V' \to U \to V$. Then $V' \to U$ is necessarily an open immersion. Namely, $V' \to f^{-1}(V') \to V'$ are monomorphisms (Schemes, Lemma 26.23.8) composing to the identity, hence isomorphisms. Thus $V'$ is isomorphic to an open of both $X$ and $Y$. In the $S$-rational maps case, the exact same argument works.
$\square$

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