Question: I prove that $ f $ has a fixed point. My question is whether the point is unique and the mapping $ f $ is continuous.

Answer: 
Paul has already shown uniqueness. Your mapping $f$ is indeed continuous. Theorem 10.1 in Munkres *Topology: A First Course* says that $f$ is continuous if, given any $x$ and a real $\epsilon > 0,$ there exists a $\delta > 0$ such that $$ d(x,y) < \delta \; \; \Longrightarrow \; \; d(f(x), f(y)) < \epsilon. $$
In this case, given $\epsilon$ we can simply take $\delta = \epsilon.$

Which leads to the shortest proof of the existence of a fixed point: the mapping
$$ g(x) = d(x, f(x)) $$
is also continuous, and is real valued. As the space is compact, $g$ achieves its minimum, at some $x\_0.$ 

If we assume $g(x\_0) > 0,$ then calculate $$ g(f(x\_0)) = d(f(x\_0), f(f(x\_0)) ) < d(x\_0, f(x\_0)) = g(x\_0), $$ which is a contradiction of the minimality of $g(x\_0).$ So, in fact, $g(x\_0) = 0$ and so $f(x\_0) = x\_0$ shows the fixed point.
Suppose $x\_0$, $x\_1$ are fixed points of $f$, i.e. $f(x\_0)=x\_0$ and $f(x\_1)=x\_1$. If $x\_0\neq x\_1$, then we can use the the inequality $d(f(x),f(y))<d(x,y)$ with $x=x\_0$ and $y=x\_1$, we have $$d(x\_0,x\_1)=d(f(x\_0),f(x\_1))<d(x\_0,x\_1),$$ which is a contradiction. Therefore, we must have $x\_0=x\_1$, i.e. there exists a unique fixed point of $f$.