Given a separated scheme morphism $X\to Y$, and a morphism $Z \to Y$, Hartshorne proves that the extension $X\times_YZ \to Z$ is also separated, as long as the schemes involved are Noetherian. The valuative criterion he uses to prove it depends on this assumption.
Is it true for general $X, Y, Z$?
$X\to Y$ separated means that the diagonal map $X\to X\times_YX$ is a closed immersion. I have to use this somehow to prove that $$ \Delta:X\times_YZ \to (X\times_YZ)\times_Z(X\times_YZ) $$ is a closed immersion. Can that product be simplified in any way? Is there an obvious way to see this that I'm missing?
As suggested in the comments, the proof is purely formal and only uses simple facts about closed immersions and pullbacks.
Assume $X \to Y$ is separated, i.e. $X \to X \times_Y X$ is a closed immersion. Consider a pullback square: $$\require{AMScd} \begin{CD} X' @>>> X \\ @VVV @VVV \\ Y' @>>> Y \end{CD}$$ The pullback pasting lemma says the outer rectangle of the following diagram is a pullback diagram, $$\begin{CD} X' \times_{Y'} X' @>>> X' @>>> X \\ @VVV @VVV @VVV \\ X' @>>> Y' @>>> Y \end{CD}$$ so the outer rectangle below is also a pullback diagram, $$\begin{CD} X' \times_{Y'} X' @>>> X \times_Y X @>>> X \\ @VVV @VVV @VVV \\ X' @>>> X @>>> Y \end{CD}$$ but the right half is a pullback square, so the left half is itself a pullback square. Thus, in the following diagram, $$\begin{CD} X' @>>> X \\ @VVV @VVV \\ X' \times_{Y'} X' @>>> X \times_Y X \\ @VVV @VVV \\ X' @>>> X \end{CD}$$ the outer rectangle and the bottom half are pullback diagrams, so the top half is a pullback square. Hence $X' \to X' \times_{Y'} X'$ is indeed a closed immersion.
A similar (but more tricky) proof shows that the class of separated morphisms is closed under composition.