Here is an approach that does not use the quantile function whatsoever - the only property used is that independent copies of $X$ have zero probability of being equal. (The main ingredient in my argument is conditional expectation.)
Consider the cumulative distribution function of $X$, namely
$$
F(t)=\mathbb P(X\leq t).
$$
Your random variable - which I will suggestively call $U$ instead of $Y$ - can be described by starting with two independent and identically distributed random variables $X,Z$ and considering the conditional probability
$$
U=\mathbb P(X\leq Z\mid Z).
$$
Then, for all integers $n\geq 1$, we can represent $U^n$ as follows. Let $X_1,X_2,\ldots,X_n,Z$ be independent and identically distributed. By independence,
$$
\mathbb P\bigl(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z\bigm\vert Z\bigr)=U^n,
$$
and thus by the tower property
$$
\mathbb EU^n=\mathbb P(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z)=\mathbb P\bigl(Z=\max(X_1,X_2,\ldots,X_n,Z)\bigr).
$$
Since $X_1,\ldots,X_n,Z$ are iid, each of them is equally likely to be the maximum and therefore
$$
\mathbb EU^n=\frac{1}{n+1}.
$$
Thus $U$ has the same moments as a uniformly distributed random variable on $[0,1]$. Since $U$ is supported in $[0,1]$ as well, it follows (by the uniqueness of the Hausdorff moment problem) that $U$ is uniformly distributed, as desired.