Prove the map has a fixed point

Define $f(x):=\rho(x,A(x))$; it's a continuous map. (Note $$\rho(x,Ax)\le\rho(x,y)+\rho(y,Ay)+\rho(Ay,Ax)\quad\forall x, y\in K$$ or $$\rho(x,Ax)-\rho(y,Ay)\le\rho(x,y)+\rho(Ax,Ay).$$ Reversing the roles of $x,y$ to get $$\left|\rho(x,Ax)-\rho(y,Ay)\right|\le\rho(x,y)+\rho(Ax,Ay)<2\delta \quad \text{ whenever }\rho(x,y)<\delta.$$ That is, $f$ is actually uniformly continuous.)

Let $\alpha:=\inf_{x\in K}f(x)$, then we can find $x_0\in K$ such that $\alpha=f(x_0)$, since $K$ is compact. If $\alpha>0$, then $x_0\neq Ax_0$ and $\rho(A(Ax_0),Ax_0)<\rho(Ax_0,x_0)=\alpha$, which is a contradiction. So $\alpha=0$ and $x_0$ is a fixed point. The assumption on $A$ makes it unique.

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Note that completeness wouldn't be enough in this case, for example consider $\mathbb R$ with the usual metric, and $A(x):=\sqrt{x^2+1}$. It's the major difference between $\rho(Ax,Ay)<\rho(x,y)$ for $x\neq y$ and the existence of $0<c<1$ such that for all $x,y,$: $\rho(Ax,Ay)\leq c\rho(x,y)$.

I don't have enough reputation to post a comment to reply to @андрэ 's question regarding where in the proof it is used that $f$ is a continuous function, so I'll post my answer here:

Since we are told that $K$ is a compact set. $f:K\rightarrow K$ being continuous implies that the $\mathrm{im}(f) = f(K)$ is also a compact set. We also know that compact sets are closed and bounded, which implies the existence of $\inf_{x\in K} f(x)$.

If it is possible to show that $f(K) \subseteq K$ is a closed set, then it is necessarily compact as well: A subset of a compact set is compact? However, I am not aware of how you would do this in this case without relying on continuity of $f$.