It is given that $g:C\longrightarrow[0,1]$ is continuous and surjective. Therefore, we know that $g^{-1}(V)$ is open if $V$ is open from common knowledge.
However, in the other direction, is it true that if $g^{-1}(V)$ is open, then $V$ is open?
Since the Cantor set is not open, this holds for $g^{-1}([0,1])=C$, but I can't seem to progress further than this.
Let $X$ be a compact space, $Y$ be a Hausdorff space and consider a surjective continuous map $g:X\rightarrow Y$. Suppose that $g^{-1}(V)$ is open for some $V\subseteq Y$. Then $F = X\setminus g^{-1}(V)$ is a closed subset of $X$. Hence it is compact. Thus $g(F)\subseteq Y$ is closed (because for Hausdorff spaces continuous image of a compact subset is compact and compact is closed). Now $$V = Y\setminus g\left(X\setminus g^{-1}(V)\right)= Y\setminus F$$ is open.