EDIT: Updated to reflect changes in Question. Made the argument clearer, and fixed an off by one error.

 1. $V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$.

 2. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$

 3. Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$.

 4. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$.

 5. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$. 

By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.

Therefore, $\nu(X) = 1$

PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"