If $f(z) = z$ for some $z\in \partial \mathbb{D}$, you have your fixed point. So assume that $f(z) \neq z$ for all $z\in\partial\mathbb{D}$. For $\varepsilon \geqslant 0$ consider
$$N(\varepsilon) :=\frac{1}{2\pi i}\int_{\partial\mathbb{D}} \frac{(1+\varepsilon)-f'(\zeta)}{(1+\varepsilon)\zeta - f(\zeta)}\,d\zeta.$$
For $\varepsilon > 0$, we have $\lvert f(z)\rvert < (1+\varepsilon)\lvert z\rvert$ on $\partial\mathbb{D}$, so Rouché's theorem asserts that $(1+\varepsilon)z - f(z)$ has the same number of zeros inside the unit disk as $(1+\varepsilon)z$, namely one. So we have $N(\varepsilon) = 1$ for $\varepsilon > 0$. But since $f(z) \neq z$ on $\partial\mathbb{D}$ by assumption, $N(\varepsilon)$ depends continuously on $\varepsilon$, so we also have
$$N(0) = \frac{1}{2\pi i}\int_{\partial \mathbb{D}} \frac{1-f'(z)}{z-f(z)}\,dz = 1,$$
i.e. $z-f(z)$ has exactly one zero in the unit disk, or, put differently, $f$ has exactly one fixed point in the unit disk.
By the maximum modulus principle, we have
$$ \sup_{z \in \overline{\Omega}} |f_n(z) - f_m(z)| = \sup_{z \in \partial \Omega} |f_n(z) - f_m(z)|. $$
Since $f_n$ converges uniformly on $\partial \Omega$ to some function, the sequence $(f_n)$ is uniformly Cauchy on $\partial \Omega$ so the right hand side tends to zero as $m,n \to \infty$. This implies that $(f_n)$ is uniformly Cauchy on $\overline{\Omega}$ so $f_n$ converge uniformly to some continuous function $f$ on $\overline{\Omega}$. In particular, $f_n$ also converges uniformly to $f$ on every compact subset of $\Omega$ so $f$ is holomorphic on $\Omega$.
Best Answer
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 $r_0<r<1$ 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.