For domains $D, \Omega\subseteq \mathbb{C}^n$, we say $D$ exhausts $\Omega$, if for each compact subset $K\subseteq \Omega$, there is an injective holomorhpic map $f:D\to \Omega$ such that $K\subseteq f(D)\subseteq \Omega$. I am having trouble proving the following statement:
If $D$ exhausts $\Omega$, where $\Omega$ is bounded, then there exists injective holomorphic maps $f_n:D\to \Omega$, such that for each compact subset $K\subseteq \Omega$, there is $N\in\mathbb{N}$ such that $K\subseteq f_n(D)$ for all $n\geq N.$
I feel that in order to give these $f_n$, we need to consider some particular(a sequence of) compact subsets of $\Omega$ but since structure of neither $D$ nor $\Omega$ is known, I am unable to proceed.
This is covered by Proof with compact sets which states that there exists a sequence of compact subsets $K_m \subset \Omega$ such that $K_m \subset \operatorname{int}(K_{m+1})$ for all $m$ and $\bigcup_{m=1}^\infty K_m = \Omega$. Note that this implies $\bigcup_{m=1}^\infty \operatorname{int}(K_m) = \Omega$.
Choose $f_m$ for $K_m$. Let $K \subset \Omega$ be compact. Since $K$ is covered by the open sets $\operatorname{int}(K_m)$, we find finitely many $m_1 , \ldots,m_k$ such that $K \subset \bigcup_{i=1}^k \operatorname{int}(K_{m_i})$. W.l.o.g. we may assume that $m_1 < m_1 < \ldots < m_k$. But then $K \subset \operatorname{int}(K_{m_k}) \subset K_{m_k}$.
Taking $N = m_k$, we see that for $n \ge N$ we get $$K \subset K_n \subset f_n(D) .$$