Definition
Let $G$ be open in $\mathbb{C}$ and $Y$ be a complete metric space and $\mathscr{F}\subset C(G,Y)$.
Then, $\mathscr{F}$ is normal if every sequence of functions in $\mathscr{F}$ contains a subsequence which converges uniformly on compact subsets of $G$ to a continuous function from $G$ to $Y$.
Let $T$ be the topology of compact convergence on $Y^G$ and $\tau$ be the compact-open topology on $C(G,Y)$. Since $G$ is a metric space, $G$ is compactly generated and $(C(G,Y),\tau)$ is a closed subspace of $(Y^G,T)$.
Let $\mathscr{F}\subset C(G,Y)$.
I think $\mathscr{F}$ is normal iff $\mathscr{F}$ is precompact in $(C(G,Y),\tau)$.
If $\mathscr{F}$ is precompact, then $\mathscr{F}$ is trivially normal.
However, I have a trouble with the other direction.
So assume that $\mathscr{F}$ is normal.
Since $(C(G,Y),\tau)$ is completely metrizable, it is sufficient to prove that $\overline{\mathscr{F}}$ is sequentially compact.
Let $\{f_n\}$ be a sequence in $\overline{\mathscr{F}}$. If $f_n$ lie solely in $\mathscr{F}$, then it contains a convergent subsequence. However, what if some $f_n$ lie in the boundary of $\mathscr{F}$? How do I prove that it still contains a convergent subsequence?
For each $n$, choose $g_n\in\mathcal{F}$ such that $d(g_n,f_n)<1/n$. Then the sequence $\{g_n\}$ has a convergent subsequence. It is now easy to see that the corresponding subsequence of $\{f_n\}$ also converges (to the same limit).