If a family of meromorphic functions is normal near each point in a region, then it's normal in the region Here's a question from Ahlfors Complex Analysis.
I'm worried that I have a gross misunderstanding of normal families, because it seems like there's not very much work involved in solving it. It's from Secton 5.5.5 "Normal Families".

If the family $\mathfrak{F}$ of analytic (or meromorphic) functions is not normal in (a region) $\Omega$, show that there exists a point $z_0$ such that $\mathfrak{F}$ is not normal in any neighborhood of $z_0$. Hint: A compactness argument.

Question: Is the following correct? 
I thought I'd try to show the contrapositive:
If a family $\mathfrak{F}$ of analytic/meromorphic functions on a region $\Omega$ is such that for every $z_0$ in the region there is a neighborhood $N(z_0)$ on which $\mathfrak{F}$ is normal, then the family is normal on $\Omega$ itself.
If Ahlfors means normal in the classical sense, then I don't see what work is involved, because it seems like you just apply Marty's Theorem: The expressions
$$
\rho(f) = 2\frac{|f'|}{1 + |f|^2} 
$$
are locally bounded in each $N(z)$, so are locally bounded in $\Omega$. By Marty's Theorem the family is normal in the whole region. This doesn't even use his hint.
Alternatively, if you want to use his compactness hint, I guess you can just as easily show equicontinuity on compacta & use Arzela-Ascoli: For any compact $E$ in the region, cover it with finitely many open discs $B_{\delta_i/4}(z_i)$ such that the family is normal on each $B_{\delta_i}(z_i)$. It's equicontinuous on each closed disc $\overline{B}_{\delta_i/2}(z_i)$. Now do the usual thing: Given $\epsilon$ find $r_i$ for each disc, and take $r = \min\{\delta_1/4, \dots, \delta_N/4, r_1, \dots, r_N \}$, etc...
 A: From the definition of normality as "every sequence of functions  contains a subsequence which converges uniformly on compact subsets in the spherical metric" (I don't remember if this is the one in the book), I would argue as follows. Let $K\subset \Omega$ be a compact set. Cover it by disks $D_i$ such that the family $\mathcal F$ is normal in $2D_i$. Choose a finite subcover by $D_{i_1},\dots, D_{i_n}$. Given any sequence in the family, choose a subsequence that converges uniformly on $D_{i_1}$, from it choose a subsubsequence that converges uniformly on $D_{i_2}$, etc. The result is a subsequence converging uniformly on $K$.
Okay, so far this subsequence depends on $K$; we need one that converges uniformly on every compact subset of $\Omega$. Use the standard device of exhaustion: $\Omega = \bigcup_{n=1}^\infty \Omega_n$ where  $\overline{\Omega_n} \subset \Omega_{n+1}$ for all $n$. (One   construction  is to fix $x_0\in\Omega$ and let $\Omega_n$ be the connected component of   $x_0$ in the set $\{x\in\Omega: \operatorname{dist}(x,\partial \Omega)>1/n\}$.) We have a subsequence converging uniformly on $\overline{\Omega_1}$; from it choose a subsequence  converging uniformly on $\overline{\Omega_2}$, ... the process is infinite but the diagonal selection produces the desired subsequence. 
