Let $D\subset \mathbb{C}$ be bounded by $|z| = 1$ and some continuum $C$ in $|z| < 1$ such that $0 \not \in D$. The sequence $(f_n)$ of holomorphic and injective functions is defined on $D$ and every domain $f_n(D)$ is bounded by $|z| = 1$ and some continuum $C_n$ in $|z| < 1$ such that $0 \not \in f_n(D)$ (just like $D$ itself).
Now let $F$ be the limit function of such a sequence. Show that $F$ is also holomorphic, injective and that $F(D)$ is also bounded by $|z| = 1$ and some continuum $C_\infty$ in $|z|<1$ such that $0 \not \in F(D)$.
I have some hints how to do that. Obviously $F$ is also holomorphic in $D$ and I already showed using Hurwitz's theorem that $F$ is non-zero and injective. What I need to show now is that $|F(z)| = 1$ for $|z|=1$. The hint for that goes like this:
Show that the convergence of $(f_n)$ on some loop $\gamma \subset D$ around the origin implies the convergence of the analytic continuations of $(f_n)$ on the curve $\gamma^*$ which is the reflection of $\gamma$ on $|z|=1$. Then follow $|f_n(z)-f_m(z)| \rightarrow 0$ for $n,m \rightarrow \infty$ and $|z|=1$ using the Maximum principle.
I already showed that the continuations converge on $\gamma^*$ when $(f_n)$ converges on $\gamma$ using the reflection principle. But how do I show $|f_n(z)-f_m(z)| \rightarrow 0$ for $n,m \rightarrow \infty$ and $|z|=1$ using the Maximum principle? And how does this imply $|F(z)|=1$ for $|z|=1$?
$f_n$ converges uniformly on $\gamma*$ means that $f_n-f_m$ converges uniformly to zero there by the Cauchy convergence criterion, but now applying the Cauchy integral theorem to express $(f_n-f_m)(z)$ on the unit circle in terms of $(f_n-f_m)(\zeta)$ on $\gamma*$ we have uniform integral convergence as the denominator is bounded away so we get $f_n-f_m$ converges uniformly on $|z|=1$
for the second point, this means that $f_n \to F$ hence $|f_n| \to |F|$ on the unit circle too, but $|f_n(w)|=1$ for all $|w|=1$ and all $n$ since by hypothesis $|f_n(w)| \le 1$ and if for some $n,|w|=1$ we have $|f_n(w)| <1$ we can pick a $|y|<1$, $f(y)=f(w)$ and get a contradiction to injectivity near $w$ inside the unit circle and $y$ since $B_{\delta}(f(y)) \subset f_n(B_{\epsilon}(y))$ for small enough $\epsilon, \delta(\epsilon)$, while $f(B_{\alpha}(w)) \cap D$ is open and nonempty)