You have made good progress so far. Let's continue from where you left off. We have:
$$\nu(\pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_m)) < \varepsilon \cdot c \mu(Z)$$
for some $m \geq 1$.

Since $\pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_m) = \pi^{-1}(B) \cap B_m^c$, we can rewrite the above as
$$\nu(\pi^{-1}(B) \cap B_m^c) < \varepsilon \cdot c \mu(Z).$$
Now, let $E \in \mathcal{A}$. We want to show that
$$\nu\left( (\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E) \right) \ge (1-\delta) \cdot c \mu(Z \cap E)$$
for a suitable choice of $Z \subseteq B$ with $\mu(B \setminus Z) < \varepsilon$. As you correctly noted, this is equivalent to showing
$$\delta \cdot c \mu(Z \cap E) \ge \nu(\pi^{-1}(Z \cap E) \setminus B_m).$$
Since $Z \cap E \subseteq B$, we have $\pi^{-1}(Z \cap E) \subseteq \pi^{-1}(B)$. Then
$$\pi^{-1}(Z \cap E) \setminus B_m = \pi^{-1}(Z \cap E) \cap B_m^c \subseteq \pi^{-1}(B) \cap B_m^c.$$
Thus,
$$\nu(\pi^{-1}(Z \cap E) \setminus B_m) \le \nu(\pi^{-1}(B) \cap B_m^c) < \varepsilon \cdot c \mu(Z).$$
We want this to be less than or equal to $\delta \cdot c \mu(Z \cap E)$, i.e., we want
$$\varepsilon \cdot c \mu(Z) \le \delta \cdot c \mu(Z \cap E).$$
However, this is not generally true, as $\mu(Z \cap E)$ can be arbitrarily small. 

The issue arises from the choice of $Z$ being made independently of $\delta$. Instead, let's choose $Z$ in a more refined way. Recall that $\pi^{-1}(B) \cap B_n \to \pi^{-1}(B) \pmod{\nu}$. Let $A_n = \pi^{-1}(B) \setminus (\pi^{-1}(B) \cap B_n) = \pi^{-1}(B) \cap B_n^c$. Then $\nu(A_n) \to 0$ as $n \to \infty$.

Choose $m$ such that $\nu(A_m) < \delta c \mu(B)$. Now, define a function $f(x) = \nu(\pi^{-1}(x) \cap B_m^c)$. Then $\int_B f d\mu = \nu(\pi^{-1}(B) \cap B_m^c) < \delta c \mu(B)$. 
Let $Z = \{x \in B : f(x) < \delta c\}$. Then
$$\int_B f d\mu = \int_Z f d\mu + \int_{B \setminus Z} f d\mu.$$
Also,
$$\int_{B \setminus Z} f d\mu \ge \delta c \mu(B \setminus Z).$$
Thus,
$$\delta c \mu(B) > \nu(\pi^{-1}(B) \cap B_m^c) \ge \int_{B \setminus Z} f d\mu \ge \delta c \mu(B \setminus Z),$$
which implies $\mu(B) > \mu(B \setminus Z)$. So, $\mu(Z) > 0$. 

Now let $E \in \mathcal{A}$. We want to show
$$\nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) \ge (1 - \delta) c \mu(Z \cap E).$$
We have
$$\nu(\pi^{-1}(Z \cap E) \setminus B_m) = \nu(\pi^{-1}(Z \cap E) \cap B_m^c) = \int_{Z \cap E} f d\mu < \int_{Z \cap E} \delta c d\mu = \delta c \mu(Z \cap E).$$
Therefore,
\begin{align*}\nu((\pi^{-1}(Z) \cap B_m) \cap \pi^{-1}(Z \cap E)) &= \nu(\pi^{-1}(Z \cap E) \cap B_m) \\ &= \nu(\pi^{-1}(Z \cap E)) - \nu(\pi^{-1}(Z \cap E) \cap B_m^c) \\ &= c \mu(Z \cap E) - \nu(\pi^{-1}(Z \cap E) \cap B_m^c) \\ &> c \mu(Z \cap E) - \delta c \mu(Z \cap E) \\ &= (1 - \delta) c \mu(Z \cap E).\end{align*}