This is Exercise 12.4.C in the July 2023 edition of Vakil's algebraic geometry book. We are proving the following theorem (Theorem 12.4.1 in the book):
Suppose $\pi:X\to Y$ is a finitely presented dominant morphism of irreducible schemes such that $K(X)/K(Y)$ (a finitely-generated field extension) has transcendence degree $r$. Then there exists a nonempty open subset $U \subseteq Y$ such that for all $q \in U$, the fiber over $q$ has pure dimension $r$, or is empty.
We have already reduced it to the case where $X=\operatorname{Spec}A$ and $Y=\operatorname{Spec}B$. Exercise 12.4.C claims that this further reduces to showing that for some non-empty open subscheme $U\subseteq Y$, the map $\pi^{-1}(U)\to U$ factors as $$\pi^{-1}(U)\to \mathbb A^r_U\to U,$$ where the first map is finite surjective and the second map is the projection. The exercise has a hint that suggests showing that the irreducible components of the fiber over any point in $U$ has dimension $\ge r$ and $\le r$ separately (but the text uses a different notation that seems to be copied from an earlier edition that states the theorem for varieties only). I am having some difficulty with proving the first inequality.
I think I have a proof with some additional assumptions (in bold), as follows:
Passing to the reduction doesn't change the underlying topological space of the fibers, so we may assume that $A$ and $B$ are integral domains and that $B$ is a subring of $A$. By shrinking $U$ if necessary, we can assume that $U=D(f)$ is a distinguished open set and so by replacing $A$ with $A_f$ and $B$ with $B_f$, we can assume $U=Y$. Let $q=[\mathfrak q]\in Y$ and $p=[\mathfrak p]\in \pi^{-1}(q)$. If we assume that $A$ and $B$ are Noetherian, we have $$\dim A_\mathfrak{p}\le \dim B_\mathfrak{q}+\operatorname{codim}_{\pi^{-1}(q)} p$$ from Exercise 12.4.A (whose solution involves thinking about systems of parameters, which seems to need Noetherian-ness in an essential way). Thus, if we can show that for every irreducible component $Z$ of $\pi^{-1}(q)$, there is some closed point $p\in Z$ such that $\dim A_\mathfrak{p}\ge \dim B_\mathfrak{q}+r$, then we are done.
We know that $A$ is a finite extension of $B[x_1,\ldots,x_r]$ and since finite surjective morphisms are stable under base change, $A_\mathfrak{q}:=A\otimes_B B_\mathfrak{q}$ is also a finite extension of $B_\mathfrak{q}[x_1,\ldots,x_r]$, which has dimension at least $\dim B_\mathfrak{q}+r$. If $p$ is a closed point of $\pi^{-1}(q)$, then $\mathfrak pA_\mathfrak{q}$ is a maximal ideal of $A_\mathfrak{q}$ and so if $A_\mathfrak{q}$ is catenary, then $$\dim A_\mathfrak{p}=\dim A_\mathfrak{q}=\dim B_\mathfrak{q}[x_1,\ldots,x_r]\ge \dim B_\mathfrak{q}+r,$$ as needed.
When I was typing up this question, I came across the algebraic geometry textbook by Görtz and Wedhorn. It seems like that this is effectively Lemma 14.111 in the book:
Let $Y$ be an irreducible locally noetherian scheme, let $X$ be an irreducible scheme, and let $f : X \to Y$ be a dominant morphism of finite type. Denote by $\eta$ the generic point of $Y$, and let $e = \dim f^{-1}(\eta)$. Then for every point $x\in X$, all irreducible components of $f^{-1}(f(x))$ have dimension $\ge e$.
However, they only have a proof for the case where $Y$ is universally catenary and claims that the general case follows from a "limit argument" reducing to the case where $Y$ is the spectrum of a finitely generated $\mathbb Z$-algebra. Since they also stated some Noetherian-ness assumptions, I suspect that this may actually be necessary.
Is it possible to modify my proof to eliminate these additional assumptions?