Verified answer

Oops, my counterexample didn't map X to X...just a minute... =)

***

Edit: I originally wrote an incorrect counterexample. Here is a proof (perhaps not the most elegant):

Suppose there is no fixed point (we reach a contradiction). Choose x_1 in X. Define the infinite sequence x_{n+1} = f(x_n). So x_2 = f(x_1), x_3 = f(x_2), and so on. Since there are no fixed points, we know x_{n+1} is not x_n. So for all n>1:

||x_{n+1} - x_n|| = ||f(x_n) - f(x_{n-1})|| < ||x_n - x_{n-1}||

So the distances between consecutive points in the sequence get smaller and smaller. So the distances ||x_{n+1} - x_n|| converge to some non-negative number a.

Case 1: Suppose ||x_{n+1} - x_n|| converges to 0 as n goes to infinity. Since {x_1, x_2, x_3, …} is an infinite sequence of points in the closed and bounded set X, there must be a convergent subsequence {x_{n_k}} that converges to a point x* in X. Then:

||f(x*) - x*||

<= ||f(x*) - f(x_{n_k})|| + ||f(x_{n_k}) - x_{n_k}|| + ||x_{n_k} - x*||

<= ||x* - x_{n_k}|| + ||f(x_{n_k}) - x_{n_k}|| + ||x_{n_k} - x*||

= ||x* - x_{n_k}|| + ||x_{n_k + 1} - x_{n_k}|| + ||x_{n_k} - x*||

Taking a limit as k goes to infinity shows the right-hand-side goes to zero, meaning f(x*)=x*, a contradiction.

Case 2: Suppose ||x_{n+1} - x_n|| converges to a>0 as n goes to infinity. Then (x_{n+1}, x_n) is an infinite sequence of vectors in the closed and bounded set X x X, so there must be convergent subsequence (x_{n_k + 1}, x_{n_k}) that converges to a point (x*, y*) in X x X. Further, it must be that ||x_{n+1} - x_n|| >= a for all n, and ||x* - y*|| = a. So ||f(x*) - f(y*)|| < ||x*-y*|| < a. But you can get a contradiction.