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A morphism $f:X\to S$ is a separated morphism when its diagonal morphismis a closed immersion.

For the requested example, take the affine plane with doubled origin: glue two copies $U,V\cong {\mathbb{A}}_k^2$ by the identity onThe opens $U$ and $V$ are affine, while their intersection is the punctured affine plane, which is not affine. Indeed, its regular functions are still $k[x,y]$; if it were affine, the canonical map to $\mathrm{Spec}k[x,y]={\mathbb{A}}_k^2$ would be an isomorphism, contrary to the missing origin. The resulting scheme is not separated: in a separated scheme, the intersection of two affine opens is the inverse image of the closed diagonal inside their affine product and is therefore affine.

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.