Show that $q(z) \neq 0$ on the neighborhood $N(z_0,r)$ by continuity. Suppose that $g$ is analytic and never zero on $N(z_0,r)$, and that $g$ has a zero of order $m$ at $z_0$.
By the factorization theorem, we have $g(z) = (z-z_0)^m q(z)$ where $q(z_0)\neq 0$ and $q(z)$ is analytic on $N(z_0,r)$.
How do i show that $q(z) \neq 0$ on the neighborhood $N(z_0,r)$ by continuity ?

By mere continuity? You can’t. However, for all $z ∈ N(z_0,r)$, you have $g(z) = (z-z_0)^m q(z)$ as you say. So $q(z) = 0$ would imply $g(z) = 0$, which is impossible for $z ≠ z_0$. And on the other hand, you already have $q(z_0) ≠ 0$ by assumption.
(Remark: You can deduce that $q(z) ≠ 0$ on $N(r',z_0)$ for some real $r' > 0$ by mere continuity of $q$ using $q(z_0) ≠ 0$, but that $r'$ might be a lot smaller than $r$.)
