The problem is:
Let $X$ be a compact metric space and $K:X\to X$ be a mapping such that $d(K(x),K(y))<d(x,y)$ for all $x,y\in X$. Prove that $X$ has a unique fixed point. (Hint: assume that otherwise $\operatorname{glb}\{d(x,K(x)):x\in X\}$ is positive and achieved as a minimum. Then get a contradiction.)
The upper part of the hint has been done and what makes me stuck is deducing a contradiction from it. Lots of thanks to everyone who is willing to help.
Let's prove that by contradiction. Lets suppose that it doesn't have a fixed point, since X is compact we have then that the set
$$ C=\{d(x,K(x))|x \in X\} $$
has a minimum $m > 0$.
With that lets consider the point $x_0$ such that $d(x_0,K(x_0)) =m$. Then we would have that if we consider $x = x_0$ and $y = K(x_0)$, which both belong to $X$,
$$ d(K(x_0), K(K(x0))) < d(x0, K(x_0)) = m $$
Which is a contradiction since we said before that $m$ was the smallest number such that $d(x,K(x)) = m$ $\forall x \in X$.
