probability theory - Let $X$ a random variable with a strictly increasing distrubution function $F_X$. Show that $Y=F_X(X) sim hom(0,1)$ distrubution.

Let $X$ a random variable with a strictly increasing distrubution function $F_X$. Show that the random variable $Y=F_X(X)$ has a $\hom(0,1)$ distrubution.

Here is what I thought:

\begin{align}
F_Y(y)&=P(Y\leq y) \\
&=P(F_X(X)\leq y) \\

&=P(P(X\leq X)\leq y)
\end{align}

The chance that $X\leq X$ is always $1$, right ? Therefore I would say that:

$$F_Y(y)=P(1\leq y)=1_{[1,\infty)}(y)$$

But this is wrong... Why is this ?