A partial converse to the stability of separatedness under base change Let $S$ be a base scheme, and let $X/S$ be an $S$-scheme. Let $T/S$ be another scheme. Denote the base change of $X$ to $T$ by $X_T := X \times_S T$.
Now separatedness (resp. properness) are stable under base change, so if $X/S$ is separated then so is $X_T/T$.
I am searching for a (partial) converse to this. In its most general form

Suppose $X$ is a scheme over a base scheme $S$, and suppose that $T/S$ is a scheme. If $X_T$ is separated (resp. proper) over $T$, is $X$ separated (resp. proper) over $S$?

In this generality the claim is false: (and this may be an overkill example) consider $G$ the affine line with doubled origin over $S = \operatorname{Spec} k[t]$ (with $S$-group scheme structure), and its base change $G_T$ to $T = \operatorname{Spec} k(t)$. The first is not separated, but the second is a group scheme over a field, hence separated.
But can we impose sufficiently strong conditions (on $X$, $S$, and $T$) under which this holds?
An example of when this would be useful is if $k$ is a field of characteristic zero, $S = \operatorname{Spec} k$, $T = \operatorname{Spec} \mathbb{C}$, and $X$ is something reasonable (since in this case we can check hausdorfness and compactness in the analytification by GAGA).
 A: The general name for this problem is "descent". The answer to your question is affirmative: if $T\to S$ is fpqc (faithfully flat and quasi-compact) then separatedness descends along $T\to S$, so $X_T/T$ separated implies $X/S$ separated. For a source, see EGA IV2, 2.7.1(i), or these notes of Matt Emerton. (For quick lookups, the best place to check whether some property is stable under descent is the table at the end of Poonen's "Rational Points on Varieties", a copy of which is available at the author's website.)
Here's a sketch of the proof: consider the (cartesian) diagram
$$\require{AMScd}
\begin{CD}
X_T @>{\Delta}>> X_T\times_T X_T\\
@VVV @VVV \\
X @>{\Delta}>> X\times_S X
\end{CD}$$
where the horizontal maps are the diagonal morphisms. The top row is actually the base change of the bottom row along the map $T\to S$, so it's fpqc as that's preserved by arbitrary base change. The top row is also a closed immersion by assumption, so it suffices to show that closed immersions descend along fpqc maps. Checking this claim is local and amounts to the statement that if $A\to B$ is a surjective map of rings and $A\to A'$ is a faithfully flat map of rings, then $A\otimes_A A'\to B\otimes_A A'$ is also surjective. So $X\to X\times_S X$ is a closed immersion, and thus $X/S$ is separated.
