prove $f(x)=x$ has a unique solution Question:  Let $f$ be a continuous function from $\mathbb{R^2} \rightarrow \mathbb{R^2}$ such that $| f (x)− f (y)| ≤  \frac {1}{3} |x−y|$.
Prove $f(x)=x$ has a unique solution. 
My sketch: There exists $\epsilon, \delta$ such that the following holds because $f$ is continuous
 $|f(x)-f(y)| \leq \epsilon \leq \frac {1}{3} |x-y| \leq |x-y|< \delta$
Thus by the squeeze lemma, if $\delta$ goes to $0$, then $f(x)=f(y)$

Assume that $f(x)=x$ and $f(y)=y$ then
$$|f(x) - f(y)| = |x-y| \leq \frac{1}{3}|x-y|$$
which is impossible unless $x=y$. This shows that $f$ cannot have more than one fixpoint. To prove that it has one fix-points see Omnomnomnom's answer. An alternative way is to appeal directly to the Banach fixed-point theorem.
