Im looking at a proof where there are two analytic functions $f,g$ where $g$ is injective and maps a domain onto the open unit disk $\mathbb{D}$ and $f$ maps the same domain onto a domain $G$ with $\mathbb{D} \subset G$.
They are saying that there exists a sequence $d_n$ in $\mathbb{D}$ with $\lim d_n = d \in \partial \mathbb{D}$ and $\lim \ f(g^{-1}(d_n)) = g \in \partial G$.
I would like to know how they can say that such sequence does exist
Let $D$ be the domain of $f$ and of $G$. I will assume that, when you say that $f$ maps $D$ on $G$, what you mean is that $f$ maps $D$ onto $G$. I will also assume that $G\neq\mathbb C$; otherwise, the statement is trivially false, since then $\partial G=\emptyset$.
Let $h=f\circ g^{-1}$. Then $h\colon\mathbb{D}\longrightarrow G$ is an analytic function. Take $w\in\partial G$. For each $n\in\mathbb N$, let $d_n\in\mathbb D$ be such that $\bigl\lvert h(d_n)-w\bigr\rvert<\frac1n$. Then, since $\overline{\mathbb{D}}$ is compact, the sequence $(d_n)_{n\in\mathbb N}$ has a convergent subsequence and we can assume, without loss of generality, that $(d_n)_{n\in\mathbb N}$ itself converges. Let $d\in\overline{\mathbb{D}}$ be its limit. If $d\in\mathbb D$, then$$h(d)=h\left(\lim_{n\to\infty}d_n\right)=\lim_{n\to\infty}h(d_n)=w,$$which is impossible, because then $w\in G$ and so, since $G$ is open (by the open mapping theorem), $w\notin\partial G$. So, $d\in\overline{\mathbb D}\setminus\mathbb{D}=\partial\mathbb D$.