The original problem is an extension of the Hurwitz theorem into the case of multivariables, which states that given a region $D\in \mathbb{C}^n$, if the sequence of analytical functions $f_k(z),z\in D$ converges uniformly to some $f$ on any compact subset $E$ of $D$, and all $f_k(z)$ doesn't assume value $0$ on $D$, prove that $f$ should whether be constant $0$ or also doesn't assume $0$ on $D$.
By the Weierstrass Theorem in the multivariable case, $f(z)$ is still analytical on $D$. However, the Uniqueness Theorem is a little diffrent from the single variable case. In single variable case we know that the inverse image of $0$ for a nonzero analytical function can't have an accumulation point in the analytical region, but the multivariable case relents it to "$f^{-1}(0)$ can't have interior point", which makes this harder to tackle than the Hurwitz Theorem in the single variable case.
I tried several methods like the maximum modulus and can't figure it out. Now I am trying another one: First of all, $f$ is analytical on $D$. If it's constant zero function, we don't need to prove anything, and if not, by the Uniqueness Theorem $f^{-1}(0)$ has no interior point (and by continuity it is closed thus nowhere dense). If we set $E_{k,n}=\{z\in D\vert |f_k(z)|<\frac{1}{n}\}$, we know that it is equivalent to $\{z\in D\vert 0<|f_k(z)|<\frac{1}{n}\}$ since no $f_k(z)$ takes value $0$, and so by continuities of $f_k(z)$, every $E_{k,n}$ is open. Then since $f_k(z)$ converges to $f(z)$, we have $$f^{-1}(0)=\bigcap_{l=1}^\infty\bigcup_{m=1}^\infty\bigcap_{n=m}^\infty E_{n,l}$$ Therefore we have the following question:
For each $E_{k,n}$ open, can $\bigcap_{l=1}^\infty\bigcup_{m=1}^\infty\bigcap_{n=m}^\infty E_{n,l}$ be nonempty closed nowhere dense?
If this question has a negative answer, the original one is proved since $f^{-1}(0)$ must be empty, and I think this question is also somehow interesting on its own. I feel this has something to do with the Baire Category Theorem, but I don't know how to proceed. Any help or hint would be appreciated (it would be the best if someone can answer the original problem from some other perspectives).
Assume that there is a zero of $f(z_1, z_2... z_n)$, say $(a_1, a_2,... ,a_n)$. Fix $a_2, a_3,... ,a_n$, then we 've got a holomophic function of one variable, $f(z_1, a_2,... ,a_n)$ which must be zero for every $z_1$, by Hurwitz theorem.