Let $f : G \rightarrow \mathbb{D}$ be a biholomorphism (a holomorphic map with holomorphic inverse), and suppose $G$ is a bounded open subset of the plane. Let $C$ be a compact connected subset of the boundary $\partial G$ , and consider the set:
$I = \{y \in \partial \mathbb{D} : \exists (x_n)_{n=1}^{\infty} \subset G, d(x_n, C ) \rightarrow 0, y = \lim_{n} f(x_n) \}$
Is $I$ necessarily of finite connectivity? That is, does $I$ necessarily have at most finitely many connected components?
It may be worth noting the image will most certainly NOT be connected. For example, take $G$ to be a slit disk, and $C$ a small compact connected interval below the tip of the slit. Then the image of $C$ under the map $f$ will correspond to two connected components of the boundary of the unit disk.
A counterexample: Take $G$ as the first quadrant with all segments from $0$ to $e^{i/n}$ for all positive integers $n$ removed, and $C = \{ 0 \}$. Then $I$ consists of countably many points on $\partial \Bbb D$ and is not of finite connectivity.
Addendum: Re your comment
With $C$ as the union of the segment from $0$ to $1/2$ and all segments from $0$ to $e^{i/n}/2$ we still have that $I$ is not of finite connectivity.