I am looking at the proof of the following theorem from Stein and Shakarchi's Complex Analysis. However, in the proof, I don't understand why we can assume without loss of generality that the sequence of functions converges uniformly on all of $\Omega$. $\Omega$ here is an open set, but the assumption only holds for compact sets, so how is there no loss in generality?
If $\{f_n\}_{n=1}^\infty$ is a sequence of holomorphic functions that converges uniformly to a function $f$ in every compact subset of an open connected set $\Omega$, then the sequence of derivatives $\{f_n'\}_{n=1}^\infty$ converges uniformly to $f'$ on every compact subset of $\Omega$.
The proof proceeds as follows. Without loss of generality assume that the sequence of functions in the theorem converges uniformly on all of $\Omega$. Given $\delta>0,$ let $\Omega_\delta$ denote the subset of $\Omega$ defined by $$\Omega_\delta = \{z \in \Omega: \bar{D_\delta}(z)\subset \Omega\}.$$ In other words, $\Omega_\delta$ consists of all points in $\Omega$ which are at distance $>\delta$ from its boundary. To prove the theorem, it suffices to show that $\{f_n'\}$ converges uniformly to $f'$ on $\Omega_\delta$ for each $\delta$. This is achieved by proving the following inequality: $$\sup_{z\in \Omega_\delta} |F'(z)|\le \frac{1}{\delta} \sup_{\zeta \in \Omega} |F(\zeta)|$$ whenever $F$ is holomorphic in $\Omega,$ since it can then be applied to $F=f_n -f$ to prove the desired fact.