This question is related to
Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X, with the difference being that $F_X$ (the probability distribution function of random variable $X$) is an arbitrary continuous distribution function, not necessarily strictly increasing.
I think the proof is similar, but we have to take care of the possibility that $F_X$ may not be $1$-to-$1$. I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved. Thanks a lot!
The goal is to show $F_Y(y)=y$ for any $y \in [0,1]$. To do so, note that $F_Y(y)\triangleq\mathbb P(\{Y\le y\})$, and
$\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.$
Since $F_X$ is continuous, $F_X^{-1}([0,y])$ must be closed. So it follows that
$\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),$ which let's denote by $a$.
Therefore, $F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad$ (Q.E.D.)
Best Answer
Your solution looks fine to me.