In the text "Basic Complex Analysis" Third Edition by Jerrold E. Marsden. I'm inquiring if there's any more alternate approach's to $\text{Proposition (1)}$ and if so I'd like a hint towards a proof ?
$\text{Proposition (1)}$
If $\sum_{k}^{\infty}g_{k}(z)$ is a uniformly convergent series of continuous functions and if $z_{n} \rightarrow z$, show that
$$\lim_{m \rightarrow \infty} \sum_{n}^{\infty}g_{n}(z_{m}) = \sum_{n}^{\infty} g_{n}(z). \tag{1.1}$$
Approach I
To address $(1)$, first we let $\Gamma\subset \mathbb{C}$, such that there exists a compact set $\psi_{r}$ such that for every trivial collection $\mu$ of open subsets there exists an $r > 0$ such that, for each $\mu \in \psi_{r}$ there is a closed disc $ \overline D(z,r)$ in $\Gamma$ such that
$$ \psi_{r} \subset \overline{\bigg(\bigcup_{\mu \in \psi} D(z,r) \bigg)}. \tag{1.2} $$
After taking care to define $\Gamma$ and $\psi_{r}$, one can initially claim that
$$\lim_{m \rightarrow \infty} \prod_{n} \exp(g_{n}(z_{m})) \rightarrow \prod_{n}(g_{n}(z)).$$
To finish our game, it's wise to note that since $\psi_{r}$ is compact, and since that $\psi_{r} \subset \Gamma$ where one has closed disk contained in $\psi_{r}$, then $\prod_{n} \exp(g_{n}(z_{m}) \rightarrow \prod_{n}(g_{n}(z))$ uniformly on $\Gamma$ as $ m \rightarrow \infty$. Then $\sum_{n}g_{n}(z)$ is continuous
Approach II
To address $(1)$, first we let $\Gamma\subset \mathbb{C}$, such that there exists a compact set $\psi_{r}$ such that for every trivial collection $\mu$ of open subsets that there exists an $r > 0$ such that, for each $\mu \in \psi_{r}$ there is a closed disc $ \overline D(z,r)$ in $\Gamma$ such that
$$ \psi_{r} \subset \overline{\bigg(\bigcup_{\mu \in \psi} D(z,r). \bigg)} \tag{2.1} $$
Differentiating $(1.1)$ we conjecture that,
$$\lim_{m \rightarrow \infty} \Bigg( D_{n} \Big( \sum_{n}g_{n}(z_{m})\Big) \Bigg) \rightarrow \Big( \sum_{n}g_{n}(z)\Big) \Bigg) \, \, \text{for all values of $m \in \psi$ }. \tag{2.2} $$
Employing the Cauchy Estimates we have that for each $\mu \in \psi$
$$\bigg| (\partial/\partial_{z})^{k} (g_{n_{1}}(z_{1}) - g_{n_{2}}(z_{2}) \bigg| \leq \frac{k!}{r^{k}}\sup_{|\zeta - z | \leq r } |(g_{n_{1}}(z_{})) - (g_{n_{1}}(z_{})) \tag{2.3} $$
$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \leq \frac{k!}{r^{k}}\sup_{\psi_{r} } |(g_{n_{1}}(z_{})) - (g_{n_{2}}(z_{})) |. \tag{2.4}$$
Cleaning things up it's trivial to note that
$$ \bigg| (\partial/\partial_{z})^{k} (g_{n_{1}}(z_{1}) - g_{n_{2}}(z_{2}) \bigg| \leq \frac{k!}{r^{k}}\sup_{\psi_{r} } |(g_{n_{1}}(z_{})) - (g_{n_{2}}(z_{})) |. $$
Finally it's easy to note that the RHS side of $(2.4)$ tends 0 as $g_{n_{1}}(z_{1}), g_{n_{2}}(z_{2}) \rightarrow \infty$. Thus it's safe to say with confidence that $D_{n} \sum_{n}g_{n}(z_{m}) \rightarrow \sum_{n}g_{n}(z) \, \, $ as $m \rightarrow \infty$ uniformly since $\psi_{r}$ is compact. Thus $D_{n} \sum_{n}g_{n}(z_{m})$ is uniformly Cauchy on $\psi_{r}$. Which implies $\sum_{n}g_{n}(z)$ is continuous.