Separated is not Zariski Hausdorff

A morphism is separated when its diagonal is a closed immersion into the scheme-theoretic fibre product $X{\times }_{S}X$ (Separated morphism of schemes). A topological space is Hausdorff exactly when its diagonal is closed in the ordinary product space, so the two conditions would be the same notion if the underlying space of $X{\times }_{S}X$ were the topological product of $\mathrm{\mid }X\mathrm{\mid }$ with itself. It is not, and this is the only reason the analogy stops.

Concretely, take a field $k$ and $X=\mathrm{Spec}k[t]$. The continuous map $\mathrm{\mid }\mathrm{Spec}k[x,y]\mathrm{\mid }\to \mathrm{\mid }\mathrm{Spec}k[x]\mathrm{\mid }\times \mathrm{\mid }\mathrm{Spec}k[y]\mathrm{\mid }$ on underlying spaces is surjective but not injective: the zero ideal and the prime $(x-y)$ of $k[x,y]$ both contract to $(0)$ in each variable, because a polynomial $f(x)$ lies in $(x-y)$ only when $f=0$, and $k[x]\cap (x-y)=0$ likewise for $y$. So a point of the scheme-theoretic product carries strictly more information than a pair of points, and closedness of the diagonal in $X{\times }_{S}X$ says nothing about pairs of distinct points of $\mathrm{\mid }X\mathrm{\mid }$.

Accordingly, the affine line ${\mathbb{A}}_k^1$ is separated over $k$ (Affine schemes and affine morphisms are separated), while its Zariski point space is far from Hausdorff. A nonempty open subset of $\mathrm{Spec}k[t]$ is the complement of a closed set $V(f)$ with $f\mathrm{\ne }0$, and the generic point $(0)$ lies in every such complement, so every two nonempty open subsets meet. Two distinct points of $\mathrm{\mid }{\mathbb{A}}_k^1\mathrm{\mid }$ therefore never have disjoint open neighbourhoods. Separatedness of a scheme is a statement about the diagonal as a closed subscheme, not about the point-set topology of the scheme, and the two conventions must not be substituted for one another.