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EDIT: This is for the normal case....

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"

EDIT: General Case (I think)

Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.

Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).

So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.

EDIT: This is for the normal case....

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"

Edit: attempt at general case removed due to error.

EDIT: This is for the normal case....

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"

EDIT: General Case (I think)

Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.

Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).

So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.

EDIT: This is for the normal case....

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"

EDIT: General Case (I think)

Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.

Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).

So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.

EDIT: This is for the normal case....

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"

EDIT: This is for the normal case....

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"

EDIT: General Case (I think)

Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.

Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).

So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.