Convergence of sequence of analytic functions

Given $\{f_{n}(z)\}$, a sequence of analytic functions in the upper half plane $\mathbb C^{+}$, where each $f_{n}(z)$ has continuous extension to the real line, and $|f_{n}(z)|\leq 1$ for all $z\in \mathbb C^{+}\cup \mathbb R$ (i.e., uniformly bounded). How to prove that there is a subsequence which converges to some analytic function $f$ (whatever the convergence is; uniformly or pointwise)?

(I think there is a theorem called Montel's theorem, but I'm not sure if we can apply it here directly!)

-
Use Cauchy's integral formula and the mean value inequality to show that $\{ f_n|_K \}$ is an equicontinuous family for each compact $K\subset \mathbb{C}^+$. Use Arzela-Ascoli to prove convergence on compact sets and Weierstrass' theorem to prove that the limit is analytic. – Jose27 Jun 13 '12 at 5:13
@Jose27: Is there a known proof in some books? reference? – Chelsie Jun 13 '12 at 10:37
What prevents you from applying Montel's theorem? This is the easy version of Montel: a uniformly bounded family is normal. en.wikipedia.org/wiki/Montel's_theorem – user31373 Jun 13 '12 at 15:00
@Leonid: So you mean we can consider $\mathbb C^{+}$ as the open subset of $\mathbb C$? and then apply the theorem directly! – Chelsie Jun 13 '12 at 18:11
Well, $\mathbb C^+$ is an open subset of $\mathbb C$, is it not? By the way, the theorem applies just as well for open subsets of the sphere $\mathbb C\cup\{\infty\}$. – user31373 Jun 13 '12 at 18:13