I am reading Ravi Vakil's notes "Foundation of Algebraic Geometry" and in Proposition 6.5.5., it states the following: Suppose $X$ and $Y$ are reduced schemes. Then $X$ and $Y$ are birational if and only if there is a dense open subscheme $U$ of $X$ and a dense open subscheme$V$ of $Y$ such that $U$ and $V$ are isomorphic.
It has a short proof and I am not sure about two things from the "only if " direction, and I would appreciate any clarification.
The proof goes as follows. First he finds dense open subschemes $X_1 \subseteq X$ and $Y_1 \subseteq Y$ along with $F: X_1 \rightarrow X$ and $G: Y_1 \rightarrow Y$, whose composition in either order is the identity where it is defined. Then he defines for $n>1$, $X_{n+1} = F^{-1}(Y_n)$ and $Y_{n+1} = G^{-1}(X_n)$. In this case, we have $G \circ F |_{X_2} = id_{X_2}$ and $F \circ G |_{Y_2} = id_{Y_2}.$ Let $X_{\infty} = \cap_{n \geq 1} X_{n}$ and $Y_{\infty} = \cap_{n \geq 1} Y_{n}$. He claims that $X_2 = X_{\infty}$ and $Y_2 = Y_{\infty}$, from which it follows that $F$ and $G$ define maps between $X_2$ and $Y_2$. These are inverse maps and we are done.
The two things I don't get is:
How is $X_2 = X_{\infty}$? (and similarly for $Y_2 = Y_{\infty}$ )
Where in the proof are we using the fact that this is a reduced scheme?
Thank you!