$g:G \rightarrow G'$ is a holomorphic function of a region $G$ onto a region (open connected) $G'$. The zero set $N(g')$ of derivative of $g$ is discrete and closed in $G$. $g$ is open mapping implies the set, $$ M:= \{b \in G' \, : \, g^{-1}(b) \subseteq N(g') \}$$ is discrete and closed in $G'$. (Remmert p286).
I don't understand why openning mapping implies the result.
Closed set: Let $g(x)=z \in G' \setminus g(M)$. Then exists $x$ such that $g'(x) \not= 0$. So there exists an nhood $U_x$ of $x$, such that for all $y \in U_x$, $g'(y) \not= 0$ by continuity. As $g$ is not locally constant, $g(U_x)$ is an open set containing $z$, and $g(U_x) \cap g(M) = \emptyset$ (if $w \in g(U_x) \cap g(M)$, then $g^{-1}(w) \not \subseteq N(g')$).
The zero set of any non-constant holomorphic function on a region $G$ is discrete, i.e. all zeros are isolated. If $b \in M$, take some $z \in g^{-1}(b)$. Then $g'(z) = 0$. If $D$ is a sufficiently small open disk around $z$, $g'(z') \ne 0$ for all $z' \in D \backslash \{z\}$. Now the Open Mapping Theorem says $g(D)$ is an open neighbourhood of $g(z) = b$, and for any $b' \in g(D) \backslash \{b\}$ we have $b' = g(z')$ for some $z' \in D \backslash \{z\}$, and $g'(z') \ne 0$ so $b' \notin M$.