See UPD part in question, answer below does not make much sense.
I think I came up with a counterexample, though it is kind of intuitive then rigorous. Consider any reasonable $F(x_0, x_1)$ on $X^2$, then for each number $a$ on $[0;1]$ there must be some "large" subset $H_a \subset X^2$ such that $F(x) = a \ \ \forall x \in H_a$, because "whatever positive direction you go from any point with $F(x_s) < a$, you will eventually hit $F(.) = a$ because you should eventually hit $F=1$". Therefore, if we take that set $H_a$ and do any type of perturbation of points there, distribution of $F(X)$ will not change. For example, choose some single line $l\in X^2$ and put all points in $H_a$ so that they lie on this line for each value of $a\in[0;1]$ such that $H_a$ intersects with $l$, we whould get a different distribution, but $F(X)=U[0;1]$ would still hold.
I am not sure about "repeat this process for each $a$" part, because $[0;1]$ is continuum and each $H_a$ has zero measure.