Suppose $X$ and $Y$ are integral schemes. If $X$ and $Y$ are birational then there is a dense open subscheme $U$ of $X$ and a dense open subscheme $V$ of $Y$ such that $U \cong V$.
My Proof: Since $X$ and $Y$ are birational, we can find dense open subsets $X_1 \subset X$, $Y_1 \subset Y$ and morphisms $F: X_1 \to Y$, $G: Y_1 \to X$ such that $G \circ F|_{F^{-1}(Y_1)} \sim \mathrm{Id}_X$ and $F\circ G|_{G^{-1}(X_1)} \sim \mathrm{Id}_Y$.
By definition of rational map, this means that we can find dense open subsets $U \subset F^{-1}(Y_1) \cap X=F^{-1}(Y_1)$ and $V \subset G^{-1}(X_1) \cap Y=G^{-1}(X_1)$ such that $G\circ F|_U = \mathrm{Id}_U$ and $F \circ G|_V = \mathrm{Id}_V$.
Therefore, $U \cong V$.
It seems to me we are done, but I see that some sources (such as the stacks project and Vakil's FOAG) have more difficult proofs.
Is there something wrong with my proof, or something I am missing?