If analytic functions $f_n$ converge uniformly and the limit is injective, then almost all $f_n$ are injective I want to prove the following.

Let $K \subset \Bbb C$ be a compact set, and let $f_n$ be analytic functions on $K$ that converge uniformly to a one-to-one analytic function $f$. Then for all large $n$, $f_n$ is one-to-one.

The converse of this theorem, that if $f_n$ are one-to-one then f is one-to-one (or constant), is called Hurwitz's theorem, and I am hinted that it can be adapted to this problem. It is proved by assuming $f(a)=f(b)$, defining $g_n = f_n-f_n(a) \to g=f-f(a)$ that has a zero on $b$, while non of $g_n$ has a zero on $b$. Taking a small enough disk around $b$ so that $a$ is not in it, then by the uniform convergence $|g_n-g|<|g|$ for all large $n$, so by Rouché $g, g_n$ have the same number of zeros in this disk, which is a contradiction since $g$ has a zero and $g_n$ doesn't.
I'm not sure how to change this to my case. I assume by contrary that there are $f(a_n)=f(b_n)$ for all large $n$, and so by passing twice to a subsequence we can have $a_{n_k}\to a, b_{n_k}\to b$ so $f(a)=f(b)$, but as user MikeMiller pointed in chat we might have $\lim a_{n_k}=\lim b_{n_k}$ so this does not contradict univalence of $f$.
To mimic the proof of Hurwitz, I define $g_n=f_n-f(a_n)$ that vanish on $a_n, b_n$ and assume $a_{n_k} \to a$. If $a=\lim a_{n_k} \neq \lim b_{n_k}=b$ then $f(a)=f(b)$ so we also assume $a=b$. Then we want to show that $g=f-f(a)$ has two zeros in a neighborhood of $a$ by Rouche, but $g_{n_k}$ does not have two zeros in small disks around $a$ (not even one zero).
The original text writes:

 A: The statement is in general not true. Take $K=\{ x +iy \in \mathbb{C} \ : \ x\in [0,1], y\in [0,1]\} $. Take
$$ f(z)= z^2, \qquad f_n(z) = z^2-\frac{1}{n} z. $$
Clearly $f, \ f_n$ are analytic. Furthermore we have
$$ \sup_{z\in K} \vert f_n(z) - f(z) \vert = \sup_{z\in K} \frac{\vert z \vert}{n} = \frac{\sqrt{2}}{n} \rightarrow 0 $$
for $n\rightarrow \infty$. Note that $f$ is injective on $K$, but $f_n$ is not, as
$$ f_n(0)=0=f_n(1/n).$$
Added: Let $K = D(0,1)$. Let
$$ f(z)=e^{z}$$
and 
$$ f_n(z)=-\frac{2e}{n}z^n+ e^{z}.$$
We have
$$\sup_{z\in K} \vert f_n(z)-f(z)\vert = \frac{e}{n}\rightarrow 0$$
for $n\rightarrow \infty$.
All these functions are analytic. Furthermore, $f$ is injective as
$$ e^z=e^w \ \Leftrightarrow \ z -w \in 2\pi i \mathbb{Z}.$$
However, the $f_n$ are not injective.
$$ f_n'(z)=-2ez^{n-1} + e^z.$$
Note that $f_n$ maps real numbers to real numbers and
$$ f_n'(0)=1$$
and
$$ f_n'\left( 1\right)
= -2e + e = -e
<0.$$
Thus, $f_n$ is not injective even restricted to $[0, 1]$. Hence, we get a counterexample for any open disc (just translate and shrink it).
Newly added: Those two counterexamples show you exactly where your proof in your answer can fail. In the first one you have that $a=0$ is zero of order $2$ of $f-f(a)$ and thus by Hurwitz theorem $g_{n_k}$ has exactly two zeros in every open disc arround for sufficient large $k$. In the second counterexample the problem is that the extension will not converge uniformely on any open neighborhood of $D(0,1)$.
Hence, what you need, is that your extension needs to converge uniformely in an open neighborhood of your compact set and that $f'(a)\neq 0$.
Schiff has chosen $\Delta$ in such a way that $f$ is univalent in an open neighborhood of $\overline{\Delta}$. Thus, you are left to check that $(f_n)_{n\geq 1}$ also converges in an open neighboorhod of $\overline{\Delta}$.
A: I strangely think I solved this positively for any compact set. Can anyone spot an error?
Let $K$ be a compact set, $f_n$ converge uniformly to f on K, and $f$ is one-to-one on $K$. assume by contradiction that there is a sequence $a_{n_k} \neq b_{n_k}$ such that $f_{n_k}(a_{n_k})= f_{n_k}(b_{n_k})$. By passing to a subsequence we may assume $a_{n_k} \to a, b_{n_k} \to b$. Then $f(a_{n_k})=f(b_{n_k})$ so if $a \neq b$ then we are done. assume $a = b$. Note $g_{n_k}=f_{n_k}-f_{n_k}(a_{n_k})$ converges uniformly to $f-f(a)$ on $K$. By Hurwitz theorem there exists $r>0$ sufficiently small so that for every large $k$, $g_{n_k}$ has exactly one zero in the disk around $a$ of radius $r$. But for $k$ large enough $a_{n_k}, b_{n_k}$ are in this disk and are two distinct roots of $g_{n_k}$, contradiction!
