Contraction Map on Compact Normed Space 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.

Edit: Note that $f$ is continuous. (Why?) Define $$g(x)=\lVert f(x)-x\rVert$$ for all $x\in K$. Use the compactness of $K$ to show that $g$ obtains a non-negative minimum. If $f$ has no fixed point, we can then derive a contradiction. (Apologies for my earlier, bogus approach.) You can further show that $f$ has a unique fixed point, if you like.

If $K$ is in a vector space, as it seems, we can prove something a bit stronger

Let $K$ be a compact subset of a normed vector space. If $f:K\to K$ is a function such that $$\|f(x)-f(y)\|\leq \|x-y\|\qquad\forall x,y\in K,$$ then $f$ has fixed point.

Indeed, for each $n\in\Bbb N$, define $f_n:K\to K$ by $$f_n(x)=\left(1-\frac1n\right)f(x).$$ Now each $f_n$ is a contraction, so the Banach's fixed point theorem applies, so for each $n\in\Bbb N$, there exist $x_n\in K$ such that $$f_n\left(x_n\right)=x_n.$$

Consider the sequence $\left(x_n\right)_{n\in\Bbb N}$. Since $K$ is compact, it must have a convergent subsequence $\left(x_{n_k}\right)_{k\in\Bbb N}$, say $$x_{n_k}\to x.$$ Finally $$\begin{align*} f_{n_k}\left(x_{n_k}\right) &= x_{n_k} &&\forall k\in\Bbb N\\ \left(1-\frac{1}{n_k}\right)f\left(x_{n_k}\right) &= x_{n_k} &&\forall k\in\Bbb N\\ \lim_{k\to\infty} \left(1-\frac{1}{n_k}\right)f\left(x_{n_k}\right) &= \lim_{k\to\infty} x_{n_k}\\ f(x) &= x &&\text{because $f$ is continuous.} \end{align*}$$