Intuitively understanding $F_X(X)$ (r.v. as argument) When considering some cdf $F_X(x)$ — e.g. from here — I’m having a hard time trying to understand what $F_X(X)$ really means.
Expanding gives $P(X \leq X)$, which at first glance should always equal $1$; the only other possibility I can think of is $1/2$, assuming that the two $X$’s are actually different (bad notation?). But then we’d have $W = F_X(X) = 1$ (or $=1/2$) from the question linked above, which makes no sense.
So how can we understand the quantity $F_X(X)$?
 A: This is a case where you are confusing yourself with incorrect use of notation.  The function $F_X: \mathbb{R} \rightarrow [0,1]$ describes the distribution of the random variable $X$, but it does not use this random variable as an implicit or explicit argument.  From the probabilistic definition of the CDF, for all $x \in \mathbb{R}$ we can validly say that:
$$F_X(x) = \mathbb{P}(X \leqslant x)
\ \quad \quad \quad (\text{Valid equation}) \quad \ \ \ $$
However, it is not valid to bring the random variable $X$ in both as the descriptor in the probabilistic definition of the function and also as its argument value.  Doing so leads you to the erroneous equation:
$$F_X(X) = \mathbb{P}(X \leqslant X)
\quad \quad \quad (\text{Erroneous equation})$$
You are correct that $\mathbb{P}(X \leqslant X) = 1$,$^\dagger$ but you are incorrect to equate this expression to the CDF evaluated using the random variable as its input.  A better way to proceed is to note that if we let $Y = F_X(X)$ then for all $0 \leqslant y \leqslant 1$ we have:
$$\begin{align}
F_Y(y) 
&= \mathbb{P}(Y \leqslant y) \\[6pt]
&= \mathbb{P}(F_X(X) \leqslant y) \\[6pt]
&= \mathbb{P}(X \leqslant F_X^{-1}(y)) \\[6pt]
&= F_X(F_X^{-1}(y)) \\[6pt]
&= y, \\[6pt]
\end{align}$$
which is the CDF of the continuous uniform distribution on the unit interval.  (In this working I have assumed that $X$ is continuous, so that its distribution funciton is invertible.  If $X$ is not continuous then the resulting distribution is not uniform.  In this case the distribution has one or more discrete "lumps" corresponding to the discrete values of the distribution.)

$^\dagger$ Indeed, the antisymmetry property of the total order $\leqslant$ means that the statement $X \leqslant X$ is a tautology.  This is an even stronger finding than saying that $\mathbb{P}(X \leqslant X) = 1$.
A: For any random variable $X$, and any function $g$ defined on $\mathrm{Supp}(X)$ the support of $X$, you can create a new random variable $Y=g(X)$. This variable is random because $X$ is. The CDF $F_X$ maps the set $\mathrm{Supp}(X)$ to the interval $[0,1]$. Think of $F_X$ as some function. You can set $g = F_X$ in my example.
Your notation is indeed confusing. I suggest having two identically distributed variables $X$ and $\tilde{X}$ and thinking about the quantity $F_X(\tilde{X})$ which is $P(X\leq \tilde{X})$ which is a random variable with support in $[0,1]$.
For example, here is an empirical CDF $\hat{F}$ for $\mathcal{N}(1,2.2)$:

If I now sample new data from the same distribution and plot the histogram of $U=\hat{F}(X)$ I get the following:

