A continuous contraction with various fixed points. Today I was reading my notes for an analysis course, and found an this as exercise:

Show, that a continuous function in a metric space such that $d_X(f(f(x))), f(x)) < d_X(f(x), x))$ has a fixed point. Here $d_X$ is the metric in $X$

The problem I have is that I think we need the space to have another property, like completeness (which is the Banach's fixed point theorem)
Am I missing something?
I would like a hint not a proof, I promise to would write the proof 
here later.

Even assuming $X$ is complete, this is not true.  For instance, define a sequence $(x_n)$ of real numbers by $x_0=0$ and $x_{n+1}=x_n+1+2^{-n}$. Then the subspace $X=\{x_n:n\in\mathbb{N}\}$ of $\mathbb{R}$ is complete, but the map $f(x_n)=x_{n+1}$ satisfies $d(f(f(x)),f(x))<d(f(x),x)$ and has no fixed point.
The statement is true if you impose the much stronger restriction that $X$ is compact.
(Incidentally, the inequality in the problem statement actually immediately implies that $f$ does not have a fixed point, since if $x$ is a fixed point then the inequality becomes $0<0$.  This does not invalidate the statement, though, since it could still be vacuously true.  In any case, I imagine the inequality was only intended to be required when $x$ is not a fixed point.)
