Prove the map has a fixed point Assume $K$ is a compact metric space with metric $\rho$ and $A$ is a map from $K$ to $K$ such that $\rho (Ax,Ay) < \rho(x,y)$ for $x\neq y$. Prove A have a unique fixed point in $K$.
The uniqueness is easy. My problem is to show that there a exist fixed point. $K$ is compact, so every sequence has convergent subsequence. Construct a sequence ${x_n}$ by $x_{n+1}=Ax_{n}$,$\{x_n\}$ has a convergent subsequence $\{ x_{n_k}\}$, but how to show there is a fixed point using $\rho (Ax,Ay) < \rho(x,y)$?

you don't need to prove completeness or define any sequence. Define a nonnegative real function
$$     h(x)  = \rho(x,f(x) )   $$
This is continuous, so its minimum is achieved at some point $x_0.$  If $h(x_0) >0,$ we see that
$$   h(f(x_0) ) = \rho( f(x_0), f(f(x_0 )) < \rho( x_0, f(x_0)) = h(x_0)   $$
Put together,
$$  h(f(x_0) ) < h(x_0)  $$
Thus the assumption of a nonzero minimum of $h$  leads to a contradiction. Therefore the minimum is actually $0,$   so $h(x_0) = 0,$  so $f(x_0) = x_0 $
