Let $(f_n)$ be a sequence of bounded and holomorphic functions on the annulus $A=\{z\in\mathbb{C}\,|\, 1<|z|<2\}$ such that each $f_n$ can be holomorphically extended to $B_2(0) = \{ z\in\mathbb{C}\,|\, |z|<2\}$. Let $(f_n)$ converge uniformly to a holomorphic and bounded function $f$ on $A$. Show that $f$ can be holomorphically extended to $B_2(0)$.
My attempt: We find a sequence $(g_n)$ of holomorphic functions on $B_2(0)$ with $f_n = g_n|_A$ for all $n\in\mathbb{N}$. As each $f_n$ is bounded, we have $|g_n| = |f_n| \leq M$ on $A$. In addition, as holomorphic functions are continuous, we infer $m:=\max_{z\in \overline{B_{3/2}(0)}} |g_n(z)|<\infty$. Together, this yields $|g_n|\leq \max\{m, M\}$ on $B_2(0)$, i.e. each $g_n$ is bounded on $B_2(0)$. If I could now show that $(g_n)$ is Cauchy with respect to the supremum norm on $B_2(0)$ (I achieved to show compact convergence on $B_2(0)$ by using the maximum modulus principle, but according to this post, this is not equivalent to uniform convergence), there would be a holomorphic and bounded limit $g: B_2(0)\rightarrow \mathbb{C}$, as the set of holomorphic and bounded functions on a domain of $\mathbb{C}$ is Banach. But then $f_n$ converges uniformly to $f$ on $A$ and $f_n = g_n|_A$ converges uniformly to $g$ on $A$. By uniqueness of limits, $f=g$ on $A$. Thus, $g$ is the desired holomorphic extension of $f$ to $B_2(0)$. How can I show that $(g_n)$ is Cauchy?
Based on the comments by Ruy, we have using the maximum modulus principle \begin{align} \sup_{|z|\leq 1} |g_n(z)-g_m(z)| &\leq \sup_{|z|\leq 3/2} |g_n(z)-g_m(z)| \leq \sup_{|z|= 3/2} |g_n(z)-g_m(z)|\\ &= \sup_{|z|=3/2} |f_n(z)-f_m(z)|\leq \sup_{z\in A} |f_n(z)-f_m(z)|<\epsilon, \end{align} as $f_n\to f$ uniformly on $A$ and convergent sequences are Cauchy. Then \begin{equation} \sup_{z\in B_2(0)} |g_n(z)-g_m(z)|\leq \max\Bigg\{ \sup_{|z|\leq 1} |g_n(z)-g_m(z)|,\,\, \sup_{z\in A} |g_n(z)-g_m(z)|\Bigg\}<\epsilon, \end{equation} so we indeed have that $(g_n)$ is Cauchy with respect to the supremum norm on $B_2(0)$.