The question is as it says in the title. Suppose $X,Y \in \mathbb{R}, X,Y\sim F, supp(F)=[0,1]^2$, such that $\mathbb{E}(X|Y=y)$ is a non-decreasing function of $y$. I want to show the reverse must hold too. I'm pretty sure it is true and straightforward but not able to show it.
My approach: I want to show that $Cov(X|_{X\in[a,b]},Y)\geq 0$ for all $a,b, 0 \leq a \leq b\leq 1$. This means the linear regression coefficient of $Y$ over $X$ is non-negative for $X$ in any such interval $[a,b]$. Now we can divide $[0,1]$ into finer and finer grids and use the fact that the regression line is an approximation of $E(Y|X)$ in any interval, which has non-negative slope in every interval. So taking the limit of this grid we have that the $E(Y|X=x)$ function is non-decreasing for all $x \in [0,1]$.
We know that $E(X|Y=y)$ is non-decreasing means the correlation between $X$ and $Y$ is non-negative (as shown here, e.g.). But I need that this is true for any interval of $X$ (easy to see that this would hold for any interval of $Y$).
Turns out the statement is not true. Thanks to a friend of mine for this counterexample. While this doesn't exactly have full support etc., it is easy to find a full support distribution "close" to this which would be a counterexample. Consider the distribution $(y,x)=(1,2) (1,10), (2,1) (2,15) (2,20)$ equally likely. $E(y|x=1)=2$ but $E(y|x=2)=1$. My subsequently posted weaker claim here still seems to hold though (at least, no counterexamples yet). Would be glad if somebody has a look.