If $(f_n)_1^{\infty}$ converge uniformly to $f$ and have no zeroes in $U$, then $f \equiv 0$ or $f$ has no zeroes in $U$ Let $U \subset \mathbb{C}$ be open and $(f_n)_1^{\infty}$ be a sequence of holomorphic functions in $U$, such that none of them have $z_0$ with $f(z_0) = 0$ in $U$, and they converge to $f$ uniformly.
I want to show that either $f$ is never $0$ in $U$, or $f \equiv 0$.
The case $f \equiv 0$ is possible: Let $f_n(z) = \frac{z}{n}$.
Consider the other case $f \not\equiv 0$. Let $(g_n)_1^{\infty}$ with $g_n(z) = \frac{1}{f_n(z)}$. This converges uniformly to $g(z) = \frac{1}{f(z)}$. If $f$ had a zero in $U$, there would be a closed path $\gamma$ in $U$ such that
$$ \int_\gamma g(z) dz \neq 0,$$
since $g$ would have a non-removable singularity in $U$. Since none of the $g_n$ has a non-removable singularity in $U$, $$ \int_\gamma g_n(z) dz = 0 $$ for all $n \in \mathbb{N}$.
Because of uniform convergence, we have
$$ 0 = \lim_{n \rightarrow \infty} \int_\gamma g_n(z) dz = \int_\gamma g(z) dz,$$
thus $g$ cannot have a non-removable singularity in $U$ and therefore $f$ has no zeroes in $U$.
I am unsure whether this proof is correct. I am especially unsure about uniform convergence of $g_n$. Is this an easily fixable issue, or would I have to rewrite my entire proof?
 A: A priori, you are allowing $f$ to have zeroes. In this case, $g$ would have poles at those points. But then, it does not make sense to talk about $g_n \to g$ uniformly.

Here's an alternate method: If $f_n \to f$ uniformly on $U$, then $f_n' \to f'$ uniformly on compact subsets of $U$. Now suppose that $f \not\equiv 0$. Then, the zeroes of $f$ (if any) are isolated. Assume that $z_0 \in U$ is in fact a zero of $f$. Let $C$ be a circle in $U$, enclosing $z_0$, oriented clockwise such that no zero of $f$ lies on $C$. Then, by the argument principle, we see that $$\oint_C \frac{f'}{f} \neq 0.$$
But on the other hand, we have $$\oint_C \frac{f'_n}{f_n} = 0$$ for all $n \in \Bbb N$. Since $f_n/f'_n \xrightarrow{\text{unif.}} f/f'$ on (the compact set) $C$, we get the desired contradiction.

Question: Why does $f_n'/f_n$ converge uniformly to $f'/f$ on $C$?

 Use the fact that $1/f$ is bounded on the compact set $C$ and that $1/f_n$ is uniformly bounded on $C$. To see the latter, note that if $\delta := \inf_C f > 0,$ then $\|f_n\|$ is eventually greater $\delta/2$.
 Now, it is easy to see that $$\left\| \frac{f_n - f}{f_n f} \right\| \to 0.$$

