A function $f$ is increasing on the closed internal $[a,b]$ Let the function $f$ be increasing on the closed internal $[a,b]$. If $a \le f(a)$ and $f(b)\le b$, prove that:

$\exists x_0\in [a,b]$, such that $f(x_0)=x_0$.

Thanks for your help.
Note that $f$ need not be continuous.
 A: Assume otherwise. Especially, $f(a)>a, f(b)<b$. Let $u_0=a,v_0=b$.
Given an interval $[u_n,v_n]\subseteq [a,b]$ with $f(u_n)>u_n, f(v_n)<v_n$ we can find a subinterval $[u_{n+1},v_{n+1}]\subset [u_n,v_n]$ with 
$$f(u_{n+1})>u_{n+1}, f(v_{n+1})<v_{n+1}\ \text{ and }\ |v_{n+1}-u_{n+1}|=\frac12|u_n-v_n|.$$ 
To do so let $w_n=\frac{u_n+v_n}{2}$. If $f(w_n)>w_n$, let $u_{n+1}=w$, $v_{n+1}=v_n$; otherwise let $u_{n+1}=u_n$, $v_{n+1}=w_n$.
Then these nested intervals converge towards a number $x_0\in[a,b]$.
Then $u_n<f(u_n)\le f(x_0)\le f(v_n)<v_n$ for all $n$ implies $f(x_0)=x_0$.
A: The solution of that question is almost the same as if we take that $f$ is countinious. By nested intervals' method we can choose such nested $[a_n,b_n]$, that
$$f(a_n)\geq a_n, f(b_n)\leq b_n.$$
Since the intervals are nested, there exists one $c\in[a_n,b_n]$ in all intervals.
Since
$$a_n\leq c \leq b_n,$$
then
$$a_n \leq f(a_n)\leq f(c) \leq f(b_n)\leq b_n.$$
$$a_n \leq f(c) \leq b_n.$$
Hence, when $n\to\infty$
$$c\leq f(c) \leq c,$$
since $a_n\to c, b_n\to c.$
Your $x_0$ is $c$ :)
A: Hint: consider $g:x\mapsto f(x) -x$.
PS: are you sure there is no assumption on $f$, e.g. continuity?
PPS: after the question has been edited — without continuity, see comments below.
A: Let $g(x)=f(x)-x$.
(*)Prove that if $x\in[a,b]$ and $g(x)>0$ then there is an $\epsilon>0 $ such that $g(x)>0$ in $[x,x+\epsilon]$. For example for $x=a$ you could take 
$\epsilon=f(a)-a$. Prove a similar property for $x\in[a,b]$ with $g(x)<0$.
Now if $r=\sup\{x\in[a,b]:g(x)>0\}$ (or $\inf\{x\in[a,b]:g(x)<0\}$) use (*) to show that $g(r)=0$.
