A separated scheme whose point space is not Hausdorff

The affine line ${\mathbb{A}}_k^1=\mathrm{Spec}k[t]$ is affine over $\mathrm{Spec}k$, so by [F1] the structure morphism is separated. The points $(0)$ and $(t)$ of $\mathrm{Spec}k[t]$ are distinct: $(0)$ is the generic point, and the maximal ideal $(t)$ is a proper nonzero prime.

Let $U\subseteq \mathrm{Spec}k[t]$ be a nonempty open subset. Its complement is a proper closed subset, hence of the form $V(f)$ for some nonzero $f\in k[t]$; since $k[t]$ is a domain, a nonzero polynomial is not in the prime ideal $(0)$, so $(0)\mathrm{\notin }V(f)$ and therefore $(0)\in U$. Thus $(0)$ belongs to every nonempty open subset.

Let $U_1$ and $U_2$ be open neighbourhoods of the distinct points $(0)$ and $(t)$. Since $U_2$ is nonempty, step 1.2 gives $(0)\in U_2$, and $(0)\in U_1$ by definition; hence $U_1\cap U_2\ni (0)$ is nonempty.

Step 2.1 shows that no two distinct points of $\mathrm{Spec}k[t]$ have disjoint open neighbourhoods, so $\mathrm{\mid }{\mathbb{A}}_k^1\mathrm{\mid }$ is not Hausdorff, while step 1.1 shows that ${\mathbb{A}}_k^1$ is separated over $k$; the implication asserted in the statement is therefore false.