Let $\{f_n\}$ be a sequence of analytic functions in $D = \{z \in \mathbb{C}: |z| < 1\}$ with $\lim_{n\to\infty}f_n(z) = f(z)$ and $|f_n(z)| \le M$, $\forall z \in D, \forall n \in \mathbb{N}$.
Show that
$$\forall \alpha \in (0,1), {f_n}\to f \mbox{ uniformly in } D_{\alpha} = \{z \in \mathbb{C}: |z| \leq \alpha\}$$
My proof:
The set $D_\alpha$ is closed and bounded (i.e compact). Each $f_n$ is analytic in $D$, then each $f_n$ are continuous on $D$.
Since $D_{\alpha} \subset D$ then $f_n$ is bounded in $D_{\alpha}$. We have a bounded continuous function in a compact set, then it is uniformly continuous.
We must show that
$$\forall \epsilon > 0, \exists N > 0, \forall z \in D_{\alpha}: |f_n(z) - f(z)| < \epsilon, \forall n\ge N$$
Pick $\epsilon > 0$, and $0 \in D_\alpha$
$$\lim_{n\to\infty}f_n(z) = f(z) \Rightarrow \exists N_0 > 0: |f_n(0) - f(0)| < \frac{\epsilon}{3}$$
Now, since $0 \in D_{\alpha}$, and $f$ is uniformly continuous in $D_{\alpha}$, then
$$\exists N_n > 0, \forall z \in D_{\alpha}: |z - 0| < N_n, |f_n(z) - f_n(0)| < \frac{\epsilon}{3} $$
I want to show:
$$|f_n(z) - f(z) | < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$
How can I complete this proof?
It seems that there is a mistake in the line that begins with "$\exists N_1 > 0$", because each $f_n$ is uniformly continuous, so $N_1$ depends on $n$. When you wrote this line, you were actually assuming that $f_n$ are equicontinuous (which is way stronger).
However, one could notice that your proof was missing something since you did not use that the $f_n$ are analytic. If the $f_n$ are not analytic, then the convergence can be not uniform on compact sets.
I would recommend writing $f_n(x) - f_n(y)$ as an integral and using the Cauchy inequalities to try bound this difference uniformly in $n$.