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.
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 $x,y,$: $\rho(Ax,Ay)\leq c\rho(x,y)$.
Let $\Omega=[0,1]$. Fix $\varepsilon\in(0,1]$ and define, for each $x\in[0,1]$,
\begin{align*}
f(x)&\equiv(1-\varepsilon)x,\\
g(x)&\equiv(1-\varepsilon)x+\varepsilon.
\end{align*}
Clearly, $\|f-g\|_{\infty}=\varepsilon$. However, the only fixed point of $f$ is $0$ and the only fixed point of $g$ is $1$, which are as far from each other as possible.
ADDED: That said, one can establish a continuity property of sorts that is worth exploring. For some $n\in\mathbb N$, let $\Omega$ be a non-empty, convex, compact subset of $\mathbb R^n$. Let $\mathcal C$ denote the set of continuous functions mapping $\Omega$ into itself. Define a correspondence $\Phi:\mathcal C\rightrightarrows\Omega$ as $$\Phi(f)\equiv\text{set of fixed points of $f$}\quad\text{for each $f\in\mathcal C$}.$$ By Brouwer’s theorem, $\Phi(f)$ is not empty for any $f\in\mathcal C$.
Endowing $\mathcal C$ with the supremum norm $\|\cdot\|_{\infty}$ and $\Omega$ with the Euclidean norm $\|\cdot\|_n$, we can establish the following:
THEOREM: The correspondence $\Phi$ is upper hemicontinuous in the sense that if $O$ is an open subset of $\Omega$, then the “inverse image” $$\Phi^{-1}(O)\equiv\{f\in\mathcal C\,|\,\Phi(f)\subseteq O\}$$ is open in $\mathcal C$.
Proof: For the sake of contradiction, suppose that $\Phi^{-1}(O)$ is not open. Then, one can find some $f\in\Phi^{-1}(O)$ and two sequences $(f_m)_{m\in\mathbb N}$ and $(x_m)_{m\in\mathbb N}$ in $\mathcal C$ and $\Omega$, respectively, such that for each $m\in\mathbb N$,
- $\|f_m-f\|_{\infty}<1/m$;
- $x_m\in\Phi(f_m)$; but
- $x_m\in\Omega\setminus O$.
Since $\Omega\setminus O$ is compact, one can take some subsequence $(x_{m_k})_{k\in\mathbb N}$ converging to some $x\in\Omega\setminus O$. For each $k\in\mathbb N$, the following holds:
\begin{align*}
\|x-f(x)\|_n&\leq \|x-x_{m_k}\|_n+\|x_{m_k}-f_{m_k}(x_{m_k})\|_n\\
&+\|f_{m_k}(x_{m_k})-f(x_{m_k})\|_n+\|f(x_{m_k})-f(x)\|_n.
\end{align*}
The first, third, and fourth terms converge to $0$ as $k\to\infty$ because of convergence in $\Omega$, convergence in $\mathcal C$, and continuity, respectively. The second term vanishes because $x_{m_k}$ is a fixed point of $f_{m_k}$ for every $k\in\mathbb N$. It follows that $\|x-f(x)\|_n=0$, that is, $x$ is a fixed point of $f$. Since $f\in\Phi^{-1}(O)$, the conclusion is that $x\in \Phi(f)\subseteq O$, which contradicts $x\in\Omega\setminus O$. $\quad\blacksquare$
The above upper-hemicontinuity property of $\Phi$ can be given an equivalent sequential characterization as follows:
THEOREM: Let
- $(f_m)_{m\in\mathbb N}$ be a sequence in $\mathcal C$ converging to $f\in\mathcal C$; and
- $(x_m)_{m\in\mathbb N}$ a sequence in $\Omega$ converging to $x\in\Omega$; such that
- $x_m$ is a fixed point of $f_m$ for each $m\in\mathbb N$, that is, $x_m\in\Phi(f_m)$.
Then, $x$ is a fixed point of $f$, that is, $x\in\Phi(f)$.
Proof: For any $m\in\mathbb N$,
\begin{align*}
\|x-f(x)\|_n&\leq\|x-x_m\|_n+\|x_m-f_m(x_m)\|_n\\
&+\|f_m(x_m)-f(x_m)\|_n+\|f(x_m)-f(x)\|_n.
\end{align*}
Proceed as before. $\quad\blacksquare$
Best Answer
Contrary to our intuition the answer is negative: There are smooth actions of compact groups $G\times D^n\to D^n$ (even finite groups) on closed $n$-disk $D^n$ (for $n$ sufficiently large) so that the action is isometric with respect to some Riemannian metric on $D^n$ and has no point in $D^n$ fixed by the group $G$. By considering the distance function defined by the Riemannian metric, one obtains actions of compact groups of isometries of the resulting metric space, which do not have a fixed point in $D^n$.
Here is where the examples are coming from:
Given a smooth manifold $M$ (possibly with boundary) and a smooth action of a compact group $G$ on $M$, denoted $G\times M\to M$, there exists a $G$-invaraint Riemannian metric $ds^2$ on $M$. This standard fact is proven by taking an arbitrary Riemannian metric $ds_o^2$ on $M$ and averaging it under the action of the group $G$. If $G$ is finite, the averaging procedures is just $$ ds^2= \frac{1}{|G|} \sum_{g\in G} g^*(ds_o^2). $$ If $G$ is compact one replaces the sum with the integral over the Haar measure of $G$.
Thus, it remains to find compact groups acting smoothly on $D^n$ without a fixed point. The first examples of this type were constructed by Conner and Richardson, in the case when $G$ is the icosahedral group, i.e., the group $I$ of orientation preserving symmetries of the regular 3-dimensional icosahedron. Later on, Oliver constructed an example of a smooth action of $SO(3)$ on $D^8$ which does not have a fixed point (in particular, the subgroup $I<SO(3)$ does not fix a point in $D^8$). You can find a detailed description of Oliver's example, references and further information in the survey:
M.Davis, A survey of results in higher dimensions, In "The Smith Conjecture", (editors: J. W. Morgan and H. Bass), Academic Press, New York, 1984, https://people.math.osu.edu/davis.12/old_papers/survey.pdf
On the positive side, if you equip $D^n$ with a Riemannian metric of nonpositive curvature $ds^2$, so that the boundary is convex, then the isometry group of $(D^n, ds^2)$ does have a fixed point in $D^n$: This is a corollary of Cartan's Fixed Point theorem. You can find a proof of the latter for instance in Petersen's "Riemannian Geometry" book. I think, Kirill in his answer was trying to reproduce the proof of Cartan's theorem, without the nonpositive curvature assumption (which, of course, cannot be done).