Is there a morphism of schemes, say $X\to Y$, such that the underlying topological map is injective but the morphism $X\to Y$ is not separated (i.e., the diagonal embedding $\Delta\colon X\to X\times_Y X$ is not closed)? Clearly, a monomorphism of schemes is separated, but I do not know if the same holds if we only demand injectivity.
No, there is not exist!
Proof: Let $f:X\to Y$ be a morphism of schemes such that the underlying map of topological spaces is injective. By construction of fibre product of schemes, one can assume that $Y$ is an affine scheme and that $X\times_YX$ admits a covering by affine open subschemes $U\times_YV$, where $U$ and $V$ are affine open in $X$. Let $z\in X\times_YX$ with projections $x\in U,y\in V$, then these points lie over the same point $w\in Y$ (see Stacks Project); by hypothesis $f(x)=w=f(y)\Rightarrow x=y$. Without loss of generality, let $U=V$. Restricting $\Delta:X\to X\times_YX$ to closed emebedding $\Delta_U:U\to U\times_YU$ ($U$ is affine, therefore it is separated); one has that $\Delta$ is the gluing morphism of $\Delta_U$'s for $U$ runs in the set of affine open subsets of $X$, that is $\Delta$ is a closed morphism of schemes, equivalently, $X$ is a separated scheme. (Q.E.D. $\Box$)
For more details, see Bosch - Algebraic Geometry and Commutative Algebra, proposition 7.4.9 and corollary 7.4.10.