I want to prove the following theorem:
Let $\lbrace f_n \rbrace$ be a sequence of analytic functions on an open $U$, converging uniformly on every compact subset $K$ of $U$ to a function $f$. Then $f$ is holomorphic. Moreover, the sequence of derivatives $\lbrace f'_n \rbrace$ converges uniformly on every compact subset $K$, and $\lim f'_n = f'$.
The book of Serge Lang (Complex Analysis) gives the proof of the first part (that $f$ is holomorphic), and I understand this part. Now I have tried showing the second part myself, using the hint from the book, which says: "Cover the compact set with a finite number of closed discs contained in $U$, and of sufficiently small radius. Cauchy's formula expresses the derivatives $f'_n$ as an integral, and one can argue as in the previous theorem."
With this hint, I came to the following proof:
Let $K$ be a compact subset of $U$. Let $r>0$ be such that $\forall z \in K$, $\bar{D}(z,r)$ is contained in $U$. We note that $\mathcal{U} := \lbrace D(z,r/2) \ | \ z \in K \rbrace$ is an open cover of $K$: $K \subseteq \cup_{z \in K} D(z,r/2)$. Because $K$ is compact, there exists a finite subcovering, and we see that
$K \subseteq \cup_{i=1}^{m} D(z_i,r/2) \subseteq \cup_{i=1}^{m} \bar{D}(z_i,r/2) =: M$,
where $z_i \in K$ for $i=1,...,m$. Pick one of these $z_i$. We know that $f_n$ is analytic on $U$ for all $n$, and the same is true for the limit function $f$. So certainly they are both analytic on $\bar{D}(z_i,r)$. Then by a corollary of Cauchy's formula, we find that for $z \in \bar{D}(z_i,r/2)$,
$|f'_n(z) - f'(z)| \leq \frac{r}{(r/2)^2} \| f_n - f \|_{C_r} = \frac{4}{r} \| f_n - f \|_{C_r}$,
where $\| f_n - f \|_{C_r}$ is the sup norm of $f_n - f$ on the circle of radius $r$. This circle is closed and bounded, hence compact, so $f_n$ converges uniformly to $f$ here. Therefore, we find that the estimate above goes to zero for $n \to \infty$. Hence $f'_n$ converges uniformly to $f'$ on $\bar{D}(z_i,r/2)$. Because this is true for every $z_i$, $i=1,...,m$, we see that $f'_n$ converges uniformly to $f'$ on all of $M$. Because $K$ is contained in $M$, we can conclude that $f'_n$ converges uniformly to $f'$ on $K$, for every compact $K$. $\square$
This seems like a correct proof to me, but I am not sure, mostly because the hint said explicitly to cover $K$ with a finite number of closed discs, and I have nowhere used this fact (the finite subcover).
Any feedback would be greatly appreciated.