The Stacks project

Suggested alternative proof: since $Y_i\to Y$ is quasi-compact quasi-separated, so is the composite $f_i:Y_i\to Y\to X$. Hence, we can take the normalization $X_i'\to X$ of $X$ in $Y_i$. To check that $X'_1\amalg X'_2\to X$ is the normalization of $X$ in $Y$, it suffices to see that the factorization $Y_1\amalg Y_2\to X'_1\amalg X'_2\to X$ of $f$ satisfies parts (1) and (2) of Lemma 29.53.4. Part (2) can be verified using the universal property of the factorization $Y_i\to X'_i\to X$ of $f_i$. We see part (1), i.e., that $\nu:X'_1\amalg X'_2\to X$ is integral: On the one hand, $\nu$ is affine, by Schemes, Lemma 26.6.8. On the other hand, let $U=\operatorname{Spec} A\subset X$ be open affine. Denote $U_i'=\operatorname{Spec} A_i'$ to the inverse image in $X_i'$. The map $A\to A_i'$ is integral. The induced map on global sections by $\nu^{-1}(U)=U_1'\amalg U_2'\to U$ is $A\to A_1'\times A_2'=A'$, which is integral (it suffices to show that the ideals $I_1=A_1'\times\{0\}$ and $I_2=\{0\}\times A_2'$ of $A'$ are integral over $A$, for $I_1+I_2=A'$).

Going to leave as is.