In the text "Function Theory of a Complex Variable" by Robert E. Greene and Steven G. Krantz I'm having difficulty verifying my appoarch to proving to Exercise 41 on (pg.100) much of the details pertaining to the proof of $\text{Proposition}\, \, \, (1.2)$ can be seen in $\text{Lemma (1.2)-(1.3)}.$ Also what makes the question unique is the particular appoarch taken throughout $\text{Lemma (1.2)-(1.3)}.$
$(1.0)$
Let $f_{n}$ be continuous on the open set $U$ and let $f_{n} \rightarrow f \in U$ uniformly converage on compact sets then prove that $f_{n}(z_{n}) \rightarrow f(z_{o})$
$\text{Proposition}\, \, \, (1.2)$
If the $f_{n}$ are also holomorphic and if $0 < k \in \mathbb{Z}$, then prove that in $(1.3)$
$(1.3)$
$$(\partial_{z}f_{n}(z_{n}))^{k} \rightarrow (\partial_{z}f(z_{o}))^{k}.$$
To prove $(1)$, one must rely on the Cauchy Inequality's and compactness arguments much of the formal developments can be seen in $\text{Lemma (1.2)-(1.3)}$
$\text{Lemma (1.2)}$
Allow $\Psi$ be a compact subset of $U$ such that there is an $r >0$ such that, for each $z \in \Psi$, the closed disc $\overline{D(z,r)}$ is contained in $U$. Fix an r. Then the set
$$\Psi_{r}=\overline{\Bigg(\bigcup_{z \in \Psi}D(z,r) \Bigg)}$$
compact subset of $U$.
$\text{Lemma (1.3)}$
$\text{Theorem 1.3} \, \, \text{(The Cauchy estimate)}$.
Let $f: U \rightarrow \mathbb{C}$ be a holomorphic function on an Open set $U$ such that $P \in U$ and assume that the closed disc $\overline D(\zeta,r)$, $r>0$ is contained in $U$.$M=\sup_{z \in \overline D(\zeta,r)}|f(z)|$ Then for $k=1,2,3...$ we have
$$\bigg| \partial_{z^{k}}f(P) \bigg | \leq \frac{k!}{r^{k}}\sup_{z \in \overline D(\zeta,r)}|f(z)|.$$
Applying the Cauchy Inequality's our original proposition in $(1)$ now becomes for each $z \in \Psi$ the following bound
$$\left| \partial_{z^{k}} (f_{n_{1}}(z_{1})-f_{n_{2}}(z_{2})) \right|\leq \frac{k!}{r^{k}}\sup_{|\zeta -z| \leq r} |f_{n_{1}}(\zeta) - f_{n_{2}}(\zeta)| \leq \frac{k!}{r^{k}}\sup_{\zeta \in \Psi_{r}}|f_{n_{1}}(\zeta)-f_{n_{2}}(\zeta)|.$$
So for all $z \in \Psi$ we can construct the following estimate
$$\left| \partial_{z^{k}} (f_{n_{1}}(z_{1})-f_{n_{2}}(z_{2})) \right| \leq \frac{k!}{r^{k}}\sup_{\zeta \in \Psi_{r}}|f_{n_{1}}(\zeta)-f_{n_{2}}(\zeta)|$$
The rhs side goes to $0$ as $n_{1},n_{2}\rightarrow \infty$, since $\Psi_{r}$ is compact such that ${f_{n}(z_{n}})$ converges uniformly on $\Psi_{r}$ Thus $(\partial / \partial_{z} f_{n}(z_{n}))$ is uniformly Cauchy on $\Psi_{r}$
$\text{Summary}$
In summary construction of the Compact Subset $\Psi_{r}$ is to allow one at least at the global level to use compactness arguments to allow for the uniform convergence of $(\partial_{z}f_{n}(z_{n}))$.In summary at an intuitive level for the subset $\Psi_{r}$ to be compact one takes $\overline{\Bigg({\bigcup_{z \in \Psi}D(z,r)} \Bigg)}$ to from $\Psi_{r}$.Which means we have a set that is not "too large" in which we can have functions defined on $\Psi_{r}$ for one to have Cauchy criterion for $((\partial / \partial_{z}f_{n}(z_{n}) \rightarrow \partial / \partial_{z}f_{n}(z_{o}))$ uniformly on $\Psi_{r}$.