Another Analysis question: continuity and compactness

[TaylorWatts]

Let I = [0,1] be the closed unit interval. Suppose f is a continuous mapping from I to I. Prove that for one x an element of I, f(x) = x.

Proof:

Since [0,1] is compact and f is continuous, f is uniformly continuous.

This is where I'm stuck. I'm wondering if I can use the fact that since max {|x-y| = 1} if |x-y| = 1 then f(x) - f(y) < max {episolin}. This of course only occurs when WLOG x=1 y=0.

Stuck as far as the rest of it goes.

 

[Moo Of Doom]

Consider the function $g : I \to \mathbb{R}$ given by $g \left( x \right) = f \left( x \right) - x$. What happens if $g \left( x \right) = 0$ for some $x$? How can you show that this has to happen in $I$? I'm sure you must know a nice theorem that you can use here.