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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.

For the cover $\mathcal{U}=\{X_1,X_2\}$ of the affine plane with doubled origin, the overlap is the punctured affine plane $U$. Since $H^0(U,{\mathcal{O}}_U)=k[x,y]$, the Čech complex beginsThis map is surjective, and the normalized complex has no terms in degrees at least two. Consequently

The Mayer-Vietoris sequence for sheaf cohomology also gives $H^1(Y,{\mathcal{O}}_Y)=0$, but its next part givesPart b with $f=1$ shows that the group on the right is infinite-dimensional. Thus

This does not contradict the acyclic cover theorem. Although $X_1$ and $X_2$ are affine, their intersection $U$ is not acyclic: it has nonzero first structure-sheaf cohomology. Equivalently, this affine cover does not satisfy the theorem's hypotheses; the doubled-origin plane is not a semi-separated scheme.