Uniform limit of injective analytic functions is injective I'm stuck on the following problem:

Let $f_n$ be a sequence of injective analytic functions on the unit disc $D$ such that $f_n$ converges uniformly to $f$ on compact subsets of $D$.  Show that $f$ is either injective or constant.

Already $f$ is analytic as an almost uniform limit.  If $f'$ is identically zero, then $f$ is constant.  Otherwise $f'$ has isolated zeroes.  If $c$ is such an isolated zero, then $f_n'$ converges uniformly to $f'$ on a small disc near $c$.  By Hurwitz's theorem, for almost all $n$ the functions $f_n'$ should also have a zero in that small disc.  
But an analytic function is locally injective at a point if and only if its derivative at the point is nonzero.  And injective implies locally injective.  So the $f_n'$ shouldn't have any zeros in $D$.
So, we can at least conclude that $f'$ is never zero on $D$.  This shows that $f$ is locally injective.  Is there an easy way to conclude injectivity from local injectivity here?

Uniform limit of injective analytic functions is injective, if this limit IS NOT CONSTANT. 
Proof. Suppose that $f(z)=a$ at two points $z_1,z_2$. Surround these two points by a Jordan curve $\gamma$ such that $f(z)\neq a$  on $\gamma$. Then according to the argument principle
$$\frac{1}{2\pi i}\int\frac{f'(z)dz}{f(z)-a}\geq 2.$$
(The right hand side equals to the number of solutions of $f(z)=a$ inside $\gamma$,
counting multiplicity). On the other hand, $f_n(z)\neq a$ on $\gamma$ when $n$ is sufficiently large, and the similar integral with $f_n$ instead of $f$ is
$\leq 1$, because $f_n$ are injective. But this is a contradiction because
$f_n\to f$ uniformly, so integral with $f_n$ must converge to the integral with $f$.  
