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The coincidence locus of two scheme morphisms is the fibre productThe diagonal morphism ${\Delta }_Y$ is a locally closed immersion, and this property is stable under base change of a morphism of schemes, so $E\hookrightarrow X$ is locally closed. The universal property of a fibre product says that a morphism $T\to X$ factors through $E$ exactly when $f{|}_T=g{|}_T$. Thus $E$ is the largest locally closed subscheme on which they coincide. If $Y$ is separated, ${\Delta }_Y$ is a closed immersion, and its base change $E\hookrightarrow X$ is closed.

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.