The key to all this is that $f(\bar{D}) \subset D$:
Since $f(\bar{D})$ is compact, there exists $r_0>0$ such that $f(\bar{D})\subset D_{r_0}$. So for any $r_0 we have $|f(z)|=|(f(z)-z)+z|<|-z|$ on $D_r$ so by Rouché's theorem $-z$ and $f(z)-z$ have the same zeroes, which is one. Since this is valid for any $r>r_0$ the uniqueness result follows.
Since $|f(z)/z|=|f(z)|$ is continuous on $\partial D$ it has a maximum $M$, and by hypothesis $M<1$. So, assuming for the moment that $f(0)=0$, by the maximum principle we get $|f(z)|\leq M|z|$ for $z\in D$. This gives that $|f_n(z)|\leq M|f_{n-1}(z)|$ in $D$, and so $|f_n(z)|\leq M^n|z|$. Taking supremums over $\bar{D}$ and then the limit as $n\to \infty$ the result follows.
Assume now that $f(0)\neq 0$ then everything just said applies to $g=h\circ f\circ h^{-1}$ (with $h$ an appropiate automorphism of the disk), and the result follows in general.