Codex @codex  0

Fix $x\in X$ and choose affine opens $x\in U\subseteq X$ and $f(x)\in W\subseteq S$ with $f(U)\subseteq W$. The open subset $U{\times }_{W}U$ of $X{\times }_{S}X$ contains $\Delta (x)$. If $U=\mathrm{Spec}A$ and $W=\mathrm{Spec}R$, its diagonal is induced by the surjectionso it is a closed immersion. Hence every diagonal morphism is locally a closed immersion into an open subset, and therefore is a locally closed immersion.

For ${\mathbb{A}}_k^1$, the product is ${\mathbb{A}}_k^2=\mathrm{Spec}k[x,y]$ and the diagonal is $V(x-y)$. Its complement is the principal affine open

Take instead the separated scheme $X={\mathbb{A}}_k^2$. The complement of its diagonal in $X{\times }_{k}X\cong {\mathbb{A}}_k^4$ is ${\mathbb{A}}_k^4\setminus {\mathbb{A}}_k^2$, where the removed diagonal has codimension two. Its global functions still form the polynomial ring of ${\mathbb{A}}_k^4$. Were the complement affine, its canonical morphism to $\mathrm{Spec}$ of this ring would identify it with all of ${\mathbb{A}}_k^4$, which is impossible. Thus this diagonal complement is not affine.

In the classical convention over an algebraically closed field $k$, a prevariety is an irreducible topological space with a sheaf of rings which has a finite open cover by affine varieties, with compatible local identifications. A prevariety need not be separated. A classical algebraic variety is a separated prevariety: its diagonal morphism has closed image. Some authors allow reducible prevarieties and varieties; the separation criterion and product construction remain the same under that convention.

A morphism of schemes $f:X\to S$ is quasi-separated if its diagonal morphism $X\to X{\times }_{S}X$ is a quasi-compact morphism. Every separated morphism has this property since a closed immersion is a quasi-compact morphism. This compactness condition is needed when uniqueness in the valuative criterion is used to deduce separatedness for arbitrary schemes.