I have the following statement of Hurwitz's Theorem which I understand how to prove:
Let G be a region and suppose the sequence $\{f_n\}$ in $H(G)$ converges to $f$. If $f\not\equiv0$, $\bar{B}(a; R)\subset G$, and $f(z)\neq0$ for $|z-a|<R$ then there is an integer $N$ such that for $n\geq N$, $f$ and $f_n$ have the same number of zeros in $B(a; R)$.
And the following corollary:
If $\{f_n\}\subset H(G)$ converges to $f$ in $H(G)$ and each $f_n$ never vanishes on $G$ then either $f\equiv 0$ or $f$ never vanishes.
I can't understand how the theorem immediately implies the corollary? I understand that under the conditions of Hurwitz's Theorem given that $f_n$ never vanishes then $f$ never vanishes. However I can't understand what happens if $f\not\equiv0$ but $f(z_0)=0$ for some $z_0$ in $|z-a|<R$.
The statement of the theorem is basically the following:
If $f_n \to f$ uniformly on compacta $f, f_n$ are holomorphic and $f$ is not the zero function, then inside each closed ball the number of zeros that $f$ has is eventually equal to the number of zeros of the $f_n$.
To get the corollary assume that $f$ is not the zero functions and $f_n \to f$ as in the theorem.
If the $f_n$ have no zeros, then the amount of zeros of the $f_n$ in any closed ball is always zero. But this is eventually equal to the amount of zeros of $f$ in this ball, so $f$ cannot have any zeros in this ball.
Now since $G$ is a region, it is open. This means that for any point $p$ in it we have a ball $B_t(p)\subset G$. Since $\overline {B_{t/2}(p)}\subset B_t(p)$ we also have a closed ball containing $p$. Since $f$ has no zeros in the closed ball, it is not zero at $p$. Since $p$ was arbitrary in $G$ f has no zeros at all.