Flat with geometrically connected and proper fibers is proper

Question

Let $S$ be a locally Noetherian scheme.

Lemma 1. If $f: X \to S$ a proper flat morphism, then it has proper fibers, and if furthermore the generic fibers are geometrically connected all its fibers are (by semi-continuity, EGA IV.15.5.4).

There is a converse:

Lemma 2. if $f$ is flat (or even just univ. submersive), finite type separated, has proper and geometrically connected fibers then it is proper (EGA IV.15.7.10).

If $S$ is a dvr, there is an intermediate result:

Lemma 3. Let $f:X \to S$ be flat separated of finite type, S a dvr. If the special fiber is proper and the generic fiber is geometrically connected then $f$ is proper. Hence the special fiber is geometrically connected by the result above.

Proof: Indeed the proof is in Serre-Tate, Good reduction of abelian varieties, Lemma 3. By fpqc descent they reduce to $S=\mathop{Spec} \hat{A}$ the spectrum of a complete noetherian ring, and use lifting of proper components of the special fiber. (The Lemma assumes that $f$ is smooth but I think they only use that it is univ open).

So the following question is natural:

Question. if $f:X \to S$ is flat separated, of finite type with geometrically connected generic fibers. If the fiber at $s$ is proper, is $f$ proper at $s$ (that is proper in a neighborhood of $s$)?

Essentially I'd like to combine the local proper valuative criterion (of EGA IV.15.7.5) with the version https://stacks.math.columbia.edu/tag/0894 where we can restrict to a dense open to reduce to the previous case. Does this work?

I think that if the fiber is proper at $s$ and all fibers of generisations of $s$ are geometrically connected, then $f$ is indeed proper at $s$ (so fibers of generisations of $s$ are proper). Indeed using the local valuative criterion we reduce exactly to Lemma 3.

Conversely if all fibers of generisations of $s$ are propers and the generic fibers (we only need this for generic $\eta$ specializing to $s$) are geometrically connected then $f$ is proper too. Indeed we take a chain $s_0 \to s_1 \to \dots s_n=s$ and we Lemma 3 to conclude that $f$ is proper at $s_1$, so its fiber is geometrically connected, so $f$ is proper at $s_2$, and so on.

Answer 1

So I think I have the following proof: since the condition is topological, we can assume that everything is reduced. Since $f$ is separated, it suffices to prove that $f$ is proper at $y$ at each irreducible component of $X$. So we reduce to $S$ and $X$ integral and $f$ dominant. But in this case we can use EGA IV.15.7.1, which states that it suffices to check the valuative criterion for $S'$ a dvr when the generic point $\eta$ of $S'$ maps to the generic point of $X$ hence also to the generic point of $Y$, and the closed point $s$ of $S'$ maps to $Y$. But then we are reduced to Lemma 3.