I am trying to solve the problem below using Montel's Theorem:
Let $\{f_n\}$ be a sequence in $A(D)$ with $\text{Re}(f_n) > 0, |f_n(0) - i| < 0.5$ for all $n.$ Prove there is a subsequence of $\{f_n\}$ that converges uniformly on compact subsets of $D.$
Attempt: Let $K \subseteq D$ be compact. By Montel's Theorem, it suffices to show the sequence is bounded on $K.$ Let $g_n = if_n,$ then $\text{Im}(g_n) > 0$ and $|g_n(0) + 1| < 0.5,$ so $g_n : D \to H$ where $H$ is the upper half-plane. We have the biholomorphism $h : H \to D$ given by $h(z) = \frac{z-i}{z+i},$ so we can define $h_n = h \circ g_n : D \to D.$ The condition $g_n(0) \in D_{0.5}(-1)$ will translate into some sort of bound on $|h_n(0) - h(-1)| = |h_n(0) - i|.$
At this point, I considered contradiction to see what the issue could be with making the sequence unbounded. If $f_n$ is unbounded, then $g_n$ is unbounded. As $z \to \infty, h(z) \to -1,$ so we need to show $|h_n(z) - i| < \sqrt{2}$ to finish off the contradiction as $|i - (-1)| = \sqrt{2}.$ I'm sure that we can show $|h_n(0) - i| < \sqrt{2},$ but I'm not sure how to expand outside $0.$ We have $K \subseteq D_r(0) = \{z : |z| \le r\}$ for some $r < 1$ (else the closure of $K,$ which is $K$ itself, will contain a point on the unit circle), so we can assume $K = D_r(0)$ for some $r<1,$ but this doesn't help much.
Some algebra reveals $h(z) - i = (1-i)\frac{z+1}{z+i},$ so $|h(z) - i| = \sqrt{2} \frac{|z+1|}{|z+i|}$ and $|h_n(z) - i| = \sqrt{2}\frac{|g_n(z) + 1|}{|g_n(z) + i|}.$ Thus, we need to show $|g_n(z) + 1| < |g_n(z) + i|.$ But I still haven't used the condition around $0$ in a meaningful way, nor the fact $K$ is compact. How should I use these facts? Note that the condition around $0$ is necessary because $f_n(z) = n$ is an unbounded sequence.
$T(z) = (z-1)/(z+1)$ maps the right half-plane conformally onto the unit disk, so that the functions $g_n = T\circ f_n$ map the unit disk $\Bbb D$ into itself.
By Montel's theorem, $(g_n)$ has a subsequence $g_{n_k}$ which converges to a function $g$, uniformly on compact subsets of $\Bbb D$.
According to Hurwitz's theorem, the limit function $g$ does not take the value $1$ in $\Bbb D$, or is identically to $1$.
The second case can not occur, because $g_{n_k}(0)$ is contained in $T(\overline{B_{1/2}(i)})$, which is a closed disk not containing the point $w =1$.
Then $f_{n_k} = T^{-1} \circ g_{n_k} = (1+g_{n_k})/(1-g_{n_k})$ converges locally uniformly to the holomorphic function $f = (1+g)/(1-g)$.