Consider the universality theorem of the uniform distribution. One way to formulate it is the following: Let $F:\mathbb{R}\rightarrow [0,1]$ be a right continuous, increasing function. Then, if $X\sim F$ (ie. $X$ is a random variable that has CDF $F$) then $F(X)\sim Uniform(0,1)$.
While I can prove it, I am confused about the concept of plugging $X$ (a random variable) into its CDF.
$X$ is a random variable, that is, X is a mapping from the sample space $S$ to $\mathbb{R}$ (considering 1-dimensional scenario). $X\sim F$ means $F(t)=P(X\le t)$ where "$X\le t$" is an event, "$X\le t$"$=\{s\in S: X(s)\le t\}$. The domain of $F$ is $\mathbb{R}$. How could we even imagine the concept of $F(X)$? $X$ is a function, coming from the space of functions, not the domain of $F$. And what would $F(X)$ even be, $P(X\le X)$? What event is "$X\le X$"? This is so weird and confusing.
$X$ is a function from $S$ to $\mathbb R$, $F$ is a function from $\mathbb R$ to $\mathbb R$, so $F(X)$ is just a composition of functions $F(X(s))$.
For example, let $X$ be an exponential random variable with unit mean. Then $F(t)=1-e^{-t}$ for $t\geq 0$. And $F(X)=1-e^{-X}$.
More concrete, for each $s\in S$, $X(s)\in\mathbb R$ is a real number, so $$ F(X(s)) = \mathbb P\{s'\in S~:~X(s')\leq X(s)\} $$ The event $\{X\leq X\}=S$ is not appeared here.