Contraction Map on Compact Normed Space has a Fixed Point Let $K$  be a compact normed space  and $f:K\rightarrow K$ such that $$\|f(x)-f(y)\|<\|x-y\|\quad\quad\forall\,\, x, y\in K, x\neq y.$$ Prove that $f$ has a fixed point.

The sequence $K \supset f(K) \supset f(f(K)) \supset \cdots$ is a nested sequence of compact sets. Then their intersection is non-empty. 
Denote the intersection by $A$. Then since 
$$A =\cap f^{(n)}(K)$$
it is easy to show that $f(A)=A$.
Note there is a typo in the problem, the inequality can only hold for $x \neq y$.
Since $A$ is non-empty and compact, by This Question you can find $a_1,a_2 \in A$ so that
$$\|a_1-a_2\| = \operatorname{diam}(A) \,.$$
I claim that $a_1=a_2$ which shows that $A$ consists of a single point.  
Indeed, as $f(A)=A$, pick $x_1,x_2 \in A$ so that $f(x_i)=a_i$. If we assume by contradiction that $a_1 \neq a_2$ then 
$$\operatorname{diam}(A)=\|a_1-a_2\| = \| f(x_1)-f(x_2) \| < \|x_1-x_2 \|$$
but this is a contradiction, as $x_1,x_2 \in A$.
As $A$ is a single point set and $f(A)=A$ we are done.
