The question is essentially in the title: $f\colon X \rightarrow Y$ is a monomorphism of schemes that is universally closed; does this imply that $f$ is a closed immersion? Any such $f$ is quasi-compact by <https://stacks.math.columbia.edu/tag/04XU>, so one may add this assumption if one wants. 


In particular, is every integral monomorphism a closed immersion? (This would improve <https://stacks.math.columbia.edu/tag/03BB>)


If $Y$ is locally Noetherian, then the answer is positive and is Proposition 3.8 (i) in Ferrand's "Monomorphismes de schemas noetheriens". But is this perhaps true for any $Y$? Or are there known counterexamples? Of course, I do not want to assume that $f$ is of finite type ;)



Is a universally closed monomorphism a closed immersion?
There is a non-surjective epimorphism $B\to C$ where $B$ and $C$ are zero-dimensional local rings (D. Lazard, see <http://www.numdam.org/item/SAC_1967-1968__2__A8_0/>). Then $\mathrm{Spec}\,(C)\to \mathrm{Spec}\,(B)$ is a monomorphism but not a closed immersion, and it is universally closed because $B\_\mathrm{red}{\simeq}C\_\mathrm{red}$.


