What I've done so far:
Let $B=\{x\in\mathbb{R}^n : \|x\| \leq 1\}$ be the closed unit ball in $\mathbb{R}^n$ equipped with the standard Euclidean metric. Let $f \colon B \to B$ be a function such that $\|f(x)-f(y)\|\leq\|x-y\|$ for all $x,y \in B$.
I have already shown $f$ has a fixed point.
Next the question asks me to:
Consider $B$ equal to the closed unit ball in $l^2$. $B = \left\{ (x_n) \in \mathbb R^\omega : \sum_{n=1}^\infty x_n^2 \leq 1 \right\}$ with the $l^2$-norm $\|x\|= \left( \sum_{n=1}^\infty x_n^2 \right)^{1/2}$, where $x=(x_n)$.
How do I prove this case does not have a fixed point?