This is an immediate consequence of the intermediate value theorem. It says that for any value between f(a) and f(b) there is an x in [a,b] which belongs to its preimage. If f(0)=0, then there is nothing to do. So assume f(0)>0. If f(1)=1, you're done, so assume f(1) < 1. Then let F(x)=f(x)-x. Show this belongs to some [p,q] with p < 0 < q and hence must fix some x in [0,1].
