If there exist sequence such that $g(x_n)=f(x_{n+1})$, then we have $g(x_0)=f(x_0)$ for some $x_0$

If there is an $n$ such that $g(x_n)=f(x_n)$, we are done. So let us assume that $g(x_n)\ne f(x_n)$ for each $n$.

Now if $g(x_n)>f(x_n)$ and $g(x_{n+1})<f(x_{n+1})$, then the continuity of $f$ and $g$ implies that and $x$ such that $f(x)=g(x)$ exists somewhere between $x_n$ and $x_{n+1}$.

The case that $g(x_n)<f(x_n)$ and $g(x_{n+1})>f(x_{n+1})$ is basically the same.

So the only two remaining cases are:
A. $(\forall n) g(x_n)>f(x_n)$
B. $(\forall n) g(x_n)<f(x_n)$

Let us discuss the case A. (The case B is similar.)

For each $n$ we have $f(x_n)<g(x_n)=f(x_{n+1})$. Since $f$ is increasing, this implies $x_n<x_{n+1}$, i.e. the sequence $(x_n)$ is monotone.

Every monotone bounded sequence must have a limit, so there exists an $x$ such that $$\lim\limits_{n\to\infty} x_n=x.$$ Now we get, using the continuity of $f$ and $g$, that $$f(x)=\lim\limits_{n\to\infty} f(x_{n+1})=\lim\limits_{n\to\infty} g(x_n)=g(x).$$