If $f_{n}$ has a dense image, then $\bigcap (f_{1}\circ\cdots\circ f_{n})(X_{n})$ is dense

I think I found a solution but its rather technical. The following is more of a sketch than a solution with full details.

Let $x_0\in X$ and $\varepsilon>0$ be given.

In a first step, one has to construct points $y_{k,m}$ for $m,k\in\Bbb N_0$ with $k\leq m$ such that $$ \begin{aligned} y_{k,m} &\in X_k \\ y_{0,0} &= x_0 \\ f_k(y_{k,m}) &= y_{k-1,m} \quad\forall m\geq k\geq 1, \\ d_k(y_{k,m},y_{k,m+1}) &< \varepsilon 2^{-m-1} \quad\forall m\geq k\geq 0. \end{aligned} $$ (This requires continuity of these functions and density of the images, and can be done recursively in $m$. This step requires some technical details that I skipped here.) Then, one can show that $\{y_{k,m}\}_{m\geq k}$ is a Cauchy sequence for each $k$. Thus, there exist points $z_k$ with $$ \lim_{m\to\infty} y_{k,m} = z_k \quad\forall k\geq0. $$ Due to continuity of $f_k$, one can show that $f_k(z_k)=z_{k-1}$ holds. Therefore, we have $$ z_0\in F_k \quad\forall k\geq 1 \qquad\text{and therefore}\qquad z_0 \in F_\infty:=\bigcap_{k\in\Bbb N} F_k. $$ Finally, one can also show that $$d_0(z_0,x_0)<\varepsilon$$ holds. Since $\varepsilon>0$ was arbitrary, this shows that $F_\infty$ is dense in $X_0$.