Sometimes it is claimed that $\zeta$ cannot be equal to zero for Re$(z)>1$ because every term in the Euler product to which $\zeta$ absolutely converges is non-zero. This can be easily refuted with
$\dfrac{1}{\zeta(1)}=0=\left( \dfrac{1}{2} \right)\left( \dfrac{2}{3} \right)\left( \dfrac{4}{5} \right)...$
Instead, it is said that Hurwitz' theorem gives the rigorous proof that Re$(z)>1\implies\zeta(z)\neq0$. The theorem is: "Let $\{f_k\}$ be a sequence of holomorphic functions on a connected open set $G$ that converge uniformly on compact subsets of $G$ to a holomorphic function $f$ which is not constantly zero on $G$. If $f$ has a zero of order $m$ at $z_0$ then for every small enough $\rho>0$ and for sufficiently large $n\in\mathbb{N}$, $f_k$ has exactly $m$ zeros in the disk defined by $|z-z_0|<\rho$, including multiplicity. These zeros converge to $z_0$ as $k\to\infty$."
I think the proof I'm looking for depends on the corollary: "Let $G$ be a connected, open set and $\{f_n\}$ a sequence of holomorphic functions which converge uniformly on compact subsets of $G$ to a holomorphic function $f$. If each $f_n$ is nonzero everywhere in $G$, then $f$ is either identically zero or also is nowhere zero."