If $f:[0,1]\rightarrow [0,1]$ is increasing function. Suppose $0A Fixed point of $f$ is a point $r$ such that $f(r)=r$. 
I am not really convinced by this theorem holds for any function (even discontinuous functions). For example, I sketched this function:

This function has $f(0)>0, f(1)<1$, it is increasing and has no value $r$ where $f(r)=r$. Am I missing something or did the question miss a condition?
 A: The function you provided is not defined in $[0,1]$. 
As $f(0)>0$, the function $f$ will be over the line $y=x$ when $x=0$. Similarly, $f$ will be below the line $y=x$ when $x=1$. Then, the key here is to find a point where the function happens to go from over the line to below the line without touching it. Let's suppose this point is $r$. This is: 
$\forall x < r : f(x) > x$
And 
$\forall x > r : f(x) < x$
Also, as the function is defined for every point in the interval $[0,1]$ and it is always increasing, 
$\forall \eta \in [0,1] f(\eta)<f(\xi) ~\forall\xi>\eta$
So, if we assume that $f(r) > r$, then for $ r+\epsilon$, $f(r+\epsilon)<r+\epsilon$ but as the function is increasing, $f(r+\epsilon)>f(r)$. When doing $\epsilon \rightarrow 0$, we obtain that $f(r)$ is both $>r$ and $<r$. Which is a contradiction. 
Similar result is obtained supposing $f(r) < r$ with $f(r-\epsilon)>r-\epsilon$. The only option to avoid the contradiction is by selecting $r$ such as $f(r)=r$. 
Another approach could be analysing the function $g(x)= f(x)-x$ and proving that $g(r)=0$. 
