I have a random variable $X$, where $0<X<1$; and a random variable $Y$. Assume $X$ and $Y$ are uncorrelated but not independent.
If I let $Z \sim binomial(p=X)$. $Z=X$ with $p=X$ and $Z=1-X$ with $p=1-X$. Is it true that $Z \mathrel{\perp\mspace{-10mu}\perp} Y|X$?
The claim doesn't hold. Let $X$ be uniform $(0,1)$, and given $X$ let $Z$ be binomial$(1,X)$, i.e. $Z$ is a single toss of a coin with success probability $X$. You can calculate that for any interval $(a,b)$ inside $(0,1)$, $$ \textstyle P(Z=1\mid X\in(a,b)) = \frac12(a+b). $$ Now define $Y$ so that $(X,Y)$ has a 'triangle' shape: $Y=h(X)$, where $$ h(x) = \begin{cases} x, & \text{if $0<x<\frac12$} \\ 1-x, & \text{if $\frac12\le x<1$} \end{cases}\;. $$ Then $Y$ and $X$ are uncorrelated, but $$\textstyle P(Z=1\mid X<\frac12)=\frac14$$ while $$\textstyle P(Z=1\mid Y>\frac14,X<\frac12)=P(Z=1\mid X\in(\frac14,\frac12))=\frac38$$