Morphisms from a proper scheme to a separated one are proper

Since $g$ is separated, [F2] makes the diagonal ${\mathrm{\Delta }}_{Y\mathrm{/}S}:Y\to Y{\times }_{S}Y$ a closed immersion.

Let $p:X{\times }_{S}Y\to Y$ be the second projection. By [F7] the fibre product $X{\times }_{S}Y$ with its projection $p$ to $Y$ is the base change of $f:X\to S$ along $g:Y\to S$; since $f$ is proper, the AC-qualified [F8] makes $p$ proper.

Put $H=(h{\mathrm{pr}}_X,{\mathrm{pr}}_Y):X{\times }_{S}Y\to Y{\times }_{S}Y$, and let ${\mathrm{\Gamma }}_h:X\to X{\times }_{S}Y$ be the graph of $h$. By [F4] the square with top arrow ${\mathrm{\Gamma }}_h$, bottom arrow ${\mathrm{\Delta }}_{Y\mathrm{/}S}$, left arrow $h$ and right arrow $H$ is Cartesian, so ${\mathrm{\Gamma }}_h$ is the base change of ${\mathrm{\Delta }}_{Y\mathrm{/}S}$ along $H$. By step 1.1 the diagonal is a closed immersion, so the AC-qualified [F5] makes ${\mathrm{\Gamma }}_h$ a closed immersion.

By [F3] the second projection of the graph ${\mathrm{\Gamma }}_h$ is $h$, so $p\circ {\mathrm{\Gamma }}_h=h$.

By the AC-qualified [F6] the closed immersion ${\mathrm{\Gamma }}_h$ is finite, hence proper.

Thus $h$ is the composite of the proper morphism ${\mathrm{\Gamma }}_h$ of step 3.1 with the proper morphism $p$ of step 1.2; by the AC-qualified [F9] the $S$-morphism $h$ is proper.

The Axiom of Choice [F10] is assumed and is used only through the four AC-qualified suppliers [F5], [F6], [F8] and [F9], in steps 2.1, 3.1, 1.2 and 4.1; the diagonal, graph and pullback identifications of steps 1.1, 2.1 and 2.2 are choice-free. If $X=\mathrm{\varnothing }$ then ${\mathrm{\Gamma }}_h$ and $h$ are empty morphisms and the same steps apply, the cited results allowing the empty fibred and zero-ring charts; if $Y=\mathrm{\varnothing }$ then $X=\mathrm{\varnothing }$ because $h$ maps into $Y$; if $h={\mathrm{id}}_X$ then $Y=X$ and $g=f$, so the conclusion is the properness of $f$ itself. No Noetherian, reducedness or nonemptiness hypothesis is used.