For every $n$, let
$$g_n(z) = \frac{f_n(z)-1}{f_n(z)+1}.$$
Then $(g_n)$ is a sequence of holomorphic functions with values in the unit disk, hence a uniformly bounded sequence, and we know that there is a subsequence $(g_{n_k})$ that is locally uniformly convergent to a holomorphic function $h \colon \mathbb{D}\to \overline{\mathbb{D}}$.
The Möbius transformation
$$T\colon w \mapsto \frac{w-1}{w+1}$$
used to obtain the $g_n$ from the $f_n$ is a homeomorphism of the Riemann sphere, hence so is its inverse
$$T^{-1} \colon w \mapsto \frac{1+w}{1-w},$$
and the locally uniform convergence of $g_{n_k}$ to $h$ is equivalent to the locally uniform convergence of $f_{n_k}$ to $T^{-1}\circ h$.
If $h(z) \equiv 1$, then the $f_{n_k}$ converge locally uniformly to $\infty$. If $h(z) \equiv c \neq 1$, the $f_{n_k}$ converge locally uniformly to the finite constant $T^{-1}(c) = \frac{1+c}{1-c}$, and if $h$ is not constant, by the open mapping theorem its image is actually contained in $\mathbb{D}$ and the $f_{n_k}$ converge locally uniformly to the non-constant function
$$z \mapsto \frac{1+h(z)}{1-h(z)}$$
with values in the right half-plane.
Try this: without the accumulation point assumption, it is possible that subsequences of $f_n$ can converge to different limit functions. However, if these limit functions are holomorphic and agree on a set with an accumulation point, then what do you know about them?
Best Answer
In any dimension, uniform convergence of derivatives and convergence in a point of functions implies uniform convergence of functions (See Baby Rudin). In the analytic case, uniform convergence of analytic functions implies analytic limit using the theorem of Morera.