Proving flatness of a finite type morphism from flatness at closed points of closed fibers

Problem Statement

Let $f: X \to Y$ be a surjective morphism of finite type between affine Noetherian schemes, where $X = \operatorname{Spec} B$ and $Y = \operatorname{Spec} A$. Suppose that for every closed point $y \in Y$ and for every $x \in X_y$ that is closed in $X_y$, the stalk map $\mathcal{O}_{Y,y} \to \mathcal{O}_{X,x}$ is flat. Is $f$ a flat morphism?

Motivation: I had recently encountered a problem where I needed to show that if $f: X \to Y$ is a morphism of finite type, then $f$ is flat globally if for every point $y \in Y$ and for every $x \in X_y$ that is closed in $X_y$, the stalk map $\mathcal{O}_{Y,y} \to \mathcal{O}_{X,x}$ is flat. I managed to solve that problem. I was playing around with what happens when we restrict the base points $y$ to be strictly closed points of $Y$, which led to the following question.

My Attempt

Let $U$ be the set of points in $X$ such that $f$ is flat at $x$. By EGA, IV, 11.3.1, we know $U$ is open. Let $Z = X \setminus U$. Suppose by way of contradiction that $Z$ is nonempty. Then $Z$ is quasicompact, so it has a closed point $x_0$. Let $y_0 = f(x_0)$. It can be shown that $x_0$ is a closed point of $X_{y_0}$. If $y_0$ is a closed point of $Y$, we get a contradiction. Consider the closure of $y_0$. It is closed in $Y$, thus quasicompact, so it contains a closed point $y_1$. $y_1$ is also closed in $Y$. But I am not sure how to proceed from here.

I also tried to use the standard fact that a ring map $A \to B$ is flat if and only if $A_{\mathfrak{m}} \to B_{\mathfrak{m}}$ is flat for all maximal ideals $\mathfrak{m}$ of $A$, but that doesn't exactly translate nicely here.