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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*} $\Box$ This theorem is inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences.