Here's what I imagine is the minimal counterexample: set $X_4 = 0$ to be a constant and forget about it.
Write $X \sim B$ if $X$ is either $0$ or $1$ with probability $1/2$.
Now choose $X_1,X_2 \sim B$ independently. Then choose
$$X_3 \qquad \begin{cases}
= X_1 & \text{if } X_1 \neq X_2 \\
\sim B \text{ (independently)} & \text{if } X_1 = X_2
\end{cases}$$
Now the pairs are independent by construction. Now consider $X_1 + X_2$. If this sum is anything even then necessarily $X_1 = X_2$, so $X_3 \sim B$. Otherwise, the sum is odd, so $X_1 \neq X_2$ and $X_3 = X_1 \sim B$ still. Hence $X_1 + X_2$ is indeed independent of $X_3$.
However, clearly $X_3$ is not independent from $(X_1,X_2)$. In fact, $X_3$ is not independent from $X_1$ alone: $3/4$ of the time they must agree.
They are not independent. In fact, they'll be organized by their percentiles. I'm going to change our notation a little bit to try and clarify this point.
Let our underlying probability space be $([0,1],\mathcal{B}[0,1],\lambda)$, the unit interval with the Borel $\sigma$-field and the Lebesgue measure.
Now, with your question, $X$ and $Y$ really do nothing more than provide probability laws. (You introduced them as random variables, but immediately took their CDFs, so it may be best to think of $X$ and $Y$ as given distributions over $\mathbb{R}$, rather than the functions from $[0,1]$ to $\mathbb{R}$.) So, we are given two CDFs, $F_X, F_Y$.
We note that $F_X^{-1} : [0,1] \rightarrow \mathbb{R}$ is a random variable with the desired distribution. Similarly, $F_Y^{-1}$. And, furthermore, when we pick $\omega \in [0,1]$, $F_X^{-1}(\omega)$ actually evaluates to the $\omega$ percentile; e.g. if $\omega = 0.5$, $F_X^{-1}(\omega)$ and $F_Y^{-1}(\omega)$ both would be equal to the median of the distributions.
For even more concreteness, consider the trivial case where $X$ and $Y$ are uniform on $[0,1]$. In this case, $F_X^{-1}(\omega) = F_Y^{-1}(\omega) = \omega$ and they exactly equal each other. Definitely not independent!
Best Answer
$X_1\circ X_2$ is a function of $X_2$ and thus independent of $X_1$, $X_3$, $X_4$ and $X_3\circ X_4$ (the latter being a function of $X_4$). It will in general not be independent of $X_2$, except in special cases, e.g. where $X_1$ is constant.