Theorem: Suppose $(X,\mu,\mathcal{B})$ is a standard probability space.  Let $T:X\to X$ be a mixing transformation.  Given any measurable set $A$ and $\epsilon > 0$, there exists $N_{A,\epsilon}\in \mathbb{N}$ such that for all $B \in \mathcal{B}$, 
$$\#\{ n\in \mathbb{N} : |\mu(B\cap T^{-n}A)-\mu(A)\mu(B)|\geq \epsilon \} \leq N_{A,\epsilon} .$$
Proof: Suppose the theorem is not true.  Then there exists a measurable set $A$ and $\epsilon >0$ such that no finite $N_{A,\epsilon}$ exists.  Since $T$ is mixing, there exists $N_1\in \mathbb{N}$ such that for $m\geq N_1$, 
$|\mu(A\cap T^{-m}A)-\mu(A)^2| < \frac{\epsilon^2}{2}$.  
Choose $N_2$ such that $\frac{1}{N_2} < \frac{\epsilon^2}{2}$.  
Let $N = 2N_1N_2$.  Since we are assuming the theorem does not hold, there exists a measurable set $B$ such that 
$$\#\{ n\in \mathbb{N} : |\mu(B\cap T^{-n}A)-\mu(A)\mu(B)|\geq \epsilon \} \geq N .$$
Thus, there exists a sequence $n_1<n_2<\ldots < n_{N_2}$ such that 
$|n_j-n_i| \geq N_1$ for $i\neq j$ and either for all $i\neq j$, 
$$\mu(B\cap T^{-n_i}A)-\mu(A)\mu(B) \geq \epsilon ,\mbox{(1)}$$
or for all $i\neq j$, 
$$\mu(A)\mu(B)-\mu(B\cap T^{-n_i}A) \geq \epsilon .\mbox{(2)}$$
Assume the first claim above holds.  A proof for the second claim is similar.  
Note for $i\neq j$, $|n_j-n_i|\geq N_1$, either $n_j>n_i$ or $n_i>n_j$.  If $n_j > n_i$, then $n_j \geq n_i + N_1$ and we have 
\begin{align}
|\mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2| &= 
|\mu(T^{-n_i}(A\cap T^{-n_j+n_i}A)) - \mu(A)^2| \\ 
&= |\mu(A\cap T^{-n_j+n_i}A) - \mu(A)^2| < \frac{\epsilon^2}{2} . 
\end{align}
If $n_i > n_j$, then $n_i\geq n_j+N_1$ and 
we still have 
\begin{align}
|\mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2| &= 
|\mu(T^{-n_j}(T^{-n_i+n_j}A\cap A)) - \mu(A)^2| \\ 
&= |\mu(T^{-n_i+n_j}A\cap A) - \mu(A)^2| < \frac{\epsilon^2}{2} . 
\end{align}
Thus,
\begin{align*}
    \int_X \big( \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)\big)^2 d\mu &= 
    \frac{1}{N_2^2}\sum_{i,j=1}^{N_2} \big( \mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2 \big) \\ 
    &= \frac{1}{N_2^2}\sum_{i=1}^{N_2} \big( \mu(T^{-n_i}A\cap T^{-n_i}A) - \mu(A)^2 \big) \\ 
    &+ \frac{1}{N_2^2}\sum_{i\neq j} \big( \mu(T^{-n_i}A\cap T^{-n_j}A) - \mu(A)^2 \big) \\ 
    &< \frac{1}{N_2} + \frac{\epsilon^2}{2} < \epsilon^2 . 
\end{align*}
Hence, by the Cauchy-Schwarz Inequality, 
\begin{align*}
    \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)&-\mu(A)| I_B(x) d\mu \\ 
    &\leq \sqrt{\int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)|^2 d\mu} \sqrt{\int_X I_B^2(x) d\mu} \\
    &< \epsilon . 
\end{align*}
Therefore,
\begin{align*}
    \frac{1}{N_2}\sum_{i=1}^{N_2} \big( \mu(B\cap T^{-n_i}A) &-\mu(A)\mu(B)\big) \\ 
    &= \int_X \Big( \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)-\mu(A)\Big) I_B(x) d\mu \\ 
    &\leq \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x) - \mu(A)| I_B(x) d\mu \\
    &< \epsilon . 
\end{align*}
This contradicts (1).  If (2) holds, we still get 
\begin{align*}
    \mu(A)\mu(B) &- \frac{1}{N_2}\sum_{i=1}^{N_2} \mu(B\cap T^{-n_i}A) \\ 
    &= \int_X \Big( \mu(A) - \frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x)\Big) I_B(x) d\mu \\ 
    &\leq \int_X |\frac{1}{N_2}\sum_{i=1}^{N_2} I_{T^{-n_i}A}(x) - \mu(A)| I_B(x) d\mu \\
    &< \epsilon ,\ \mbox{which is a contradiction.} 
\end{align*}
$\Box$