I'm reading Qing Liu's book chapter 3.2.
He claims that if the image of diagonal morphism is a closed set, then the scheme is separated (Corollary 3.3.5) and omits the proof. It seems that he want to use exercise 3.3.1 and the previous proposition. But I'm confused how to reduce it into affine case.
Corollary 3.3.5: Let $f:X\rightarrow Y$ be a morphsim such that $\Delta(X)$ is a closed subset of $X\times_YX$. Then $X$ is separated.
Exercise 3.3.1: Let $f:X\rightarrow Y$ be a morphism. We suppose there exist open subsets $Y_i$ of $Y$ such that $X = \cup_if^{-1}(Y)$ and that the restrictions $f:f^{-1}(Y_i)\rightarrow Y_i$ are closed immersions. Then if $f(X)$ is closed in $Y$, we have $f$ is closed immersion.
Proposition 3.4: Any morphism of affine schemes is separated.
The original statements are:
Proposition 3.4. Any morphism of affine schemes $X \rightarrow Y$ is separated. In particular, any affine scheme is separated.
Proof Let $X=\operatorname{Spec} B$ and $Y=\operatorname{Spec} A .$ By construction of $X \times Y X, \Delta$ is induced by the homomorphism $\rho: B \otimes_{A} B \rightarrow B$ defined by $\rho\left(b_{1} \otimes b_{2}\right)=b_{1} b_{2} .$ It is clear that $\rho$ is surjective, and therefore $\Delta$ is a closed immersion.
Corollary 3.5. Let $f: X \rightarrow Y$ be a morphism of schemes such that $\Delta(X)$ is a closed subset of $X \times_{Y} X .$ Then $f$ is separated.
Proof This is a consequence of the proposition above and Exercise 3.1 .
3.1. Let $f: X \rightarrow Y$ be a morphism of schemes. We suppose that there exist open subsets $Y_{i}$ of $Y$ such that $X=\cup_{i} f^{-1}\left(Y_{i}\right),$ and that the restrictions$f: f^{-1}\left(Y_{i}\right) \rightarrow Y_{i}$ are closed immersions. Show that if $f(X)$ is closed in $Y,$ then $f$ is a closed immersion.
Write $X=\cup_i{U_i}$ as a reunion of affine subschemes, with $f(U_i)$ being contained in an affine subscheme $V_i$ of $Y$.
Then apply Exercise 3.3.1 to the map $\Delta$ and the open subsets of the base $U_i \times_{V_i} U_i$.