From "Mathematical Analysis" of T.M. Apostol:
6.5 Let $f$ be a real-valued function defined on $[0, 1 ]$ such that $f(0) > 0$, $f(x) \neq x$ for all $x$, and $f (x) \leq f (y)$ whenever $x \leq y$. Let $A = {x : f (x) > x}$. Prove that $sup A \in A$ and that $f(1) > 1$.
$0 \in A$ and $a \in A \implies a \leq 1$ hence $0 \leq sup A \leq 1$.
Consider $g(x)=f(x)-x$, we know that $g(0)>0$. Define $s=sup A$, if $s \notin A$ then $g(s) < 0$. If $f(x)$ is continuous then also $g(x)$ is continuous and for the intermediate value theorem there is point $p$ where $g(p)=0$, a contraddiction.
How to generilize for a monotone function $f$?
First part: There exists a sequence $(s_n)$ in $A$ increasing to $s$. We have $f(s) \geq f(s_n) >s_n$ for all $n$. Taking limits we get $f(s) \geq s$. Since $f(s) \neq s$ we get $f(s) >s$.