Problem: Let $f:[a,b]\to[a,b]$ be a function such that $|f(x)-f(y)|\leq|x-y|$ for every $x,y\in[a,b]$. Choose $x_1\in[a,b]$ and define $x_{n+1}=\frac{x_n+f(x_n)}{2}$. Prove that the sequence $\{x_n\}$ converges to a fixed point in $[a,b]$.
My approach: I know that if I can show that $x_n\to l$, $l\in[a,b]$ then clearly, $\lim_{n\to\infty}{x_{n+1}}=\lim_{n\to\infty}\frac{x_n+f(x_n)}{2}\implies 2l=l+f(l)\implies f(l)=l$ since $f$ is uniformly continuos on $[a,b]$. Now I got $$|x_{n+1}-x_n|\leq|x_2-x_1|\tag{1} $$ but I am unable to proceed any further in proving the convergence of $\{x_n\}$. Now I am thinking if there is any way to proceed from the inequality $(1)$ to prove the convergence of $\{x_n\}$ using uniform continuity of $f$ or by any other way. Anyway, please help me to prove the convergence of $\{x_n\}$.
Thank you.
(Compare 1, 2 on AoPS): The function $$ g(x) = \frac{x + f(x)}{2} $$ is monotonically increasing on $[a, b]$: For $a \le x < y \le b$ we have $$ f(x) - f(y) \le |f(x) - f(y)| \le |x-y| = y-x \\ \implies x +f(x) \le y + f(y) \\ \implies g(x) \le g(y) \, . $$
It follows that the sequence $(x_n)$ is monotonically increasing or decreasing. The sequence is also bounded, and therefore convergent.