Wow, what a beautify theorem. I never really appreciated fixed points theorems until now.

Define a function on the closed interval [0,1]

g(x)=f(x)-x

We note that since f(x) and x are continous on [0,1] so too g(x) is continous on [0,1].

We note that g(0)=f(0)-0=f(0)

And g(1)=f(1)-1

Here is the crucial steps:

f(0)=>0 by the defintion of f(x).

If f(0)=0 then the proof is complete, that is c=0.

Thus, there is not harm in assuming f(0)>0

f(1)<=1 thus, f(1)-1<=0

If f(1)-1=0 the proof is complete, that is c=1.

Thus, there is no harm in assuming f(1)-1<0

Since g(x) is continous by intermediate value theorem, [zero is between the f(0) and f(1)-1]

There is a point c in [0,1] such that, g(c)=0

But, g(c)=f(c)-c thus, f(c)=c.

Q.E.D.