The answer to my question is yes!! [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences." I now give the proof of the answer to my question, in my own words.

First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

Definition. We say that $f_n$ converges uniformly modulo re-ordering to $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$, $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

Proposition. Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$ Then $f_n$ converges uniformly modulo re-ordering to $f$.

Proof. Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$

Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by \begin{align*} f_n(B) &= \mu(A \cap T^{-n}(B)) \\ f(B) &= \mu(A)\mu(B). \end{align*}

Theorem. $f_n$ converges uniformly modulo re-ordering to $f$.

We now prove this, using the above Proposition. For each $N \in \mathbb{N}$, define the probability measure $\mu_N$ on $X^{N+1}$ to be the image measure of $\mu$ under the map $x \mapsto (x,T(x),\ldots,T^N(x))$. For each $n,N$ with $0 \leq n \leq N$, let $\pi_{N,\,n} \colon X^{N+1} \to X$ be the projection sending $(x_0,\ldots,x_N)$ to $x_n$. Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$, \begin{align*} f_n(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mathbf{1}_A \circ \pi_{N,\,N-n} \, d\mu_N \\ f(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mu(A) \, d\mu_N. \end{align*} So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives $$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{\pi_{\max J,\,\max J}^{-1}(B)} \underbrace{\mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_A \circ \pi_{\max J,\,\max J-n}}_{=: \, g_J} \, d\mu_{\max J}. $$ Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

Lemma. Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \| g_J \|_{L^2(\mu_{\max J})} \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

Proof. We have \begin{align*} \| g_J \|_{L^2}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n}) \right\|_{L^2}^2 \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J+1}} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J+1}} \mu(A)^2 - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2} - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1} + (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_{X^{\max J+1}} (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} - \mu(A)^2 \right) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square \end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]

The answer to my question is yes!! [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences." I now give the proof of the answer to my question, in my own words.

First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

Definition. We say that $f_n$ converges uniformly modulo re-ordering to $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$, $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

Proposition. Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$ Then $f_n$ converges uniformly modulo re-ordering to $f$.

Proof. Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$

Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by \begin{align*} f_n(B) &= \mu(A \cap T^{-n}(B)) \\ f(B) &= \mu(A)\mu(B). \end{align*}

Theorem. $f_n$ converges uniformly modulo re-ordering to $f$.

We now prove this, using the above Proposition. We will equip the space of $[0,1]$-valued functions on $X$ identified up to $\mu$-almost sure equality with its natural topology, namely the topology of $L^p$ convergence for any $p \in [1,\infty)$ or equivalently the topology of convergence in probability. 

Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$, \begin{align*} f_n(B) &= \int_{T^{-N}(B)} \mathbf{1}_{T^{n-N}(A)} \, d\mu \\ f(B) &= \int_{T^{-N}(B)} \mu(A) \, d\mu. \end{align*} So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives $$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{T^{-\max J}(B)} \mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \, d\mu. $$ Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

Lemma. Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_{T^{n-\max J}(A)} \to \mu(A) \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

Proof. We have \begin{align*} \| \mathrm{RHS}-\mathrm{LHS} \|_{L^2(\mu)}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_{T^{n-\max J}(A)}) \right\|_{L^2(\mu)}^2 \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X (\mu(A) - \mathbf{1}_{T^{n_1-\max J}(A)})(\mu(A) - \mathbf{1}_{T^{n_2-\max J}(A)}) \, d\mu \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_X \mu(A)^2 - \mu(A)\mathbf{1}_{T^{n_2-\max J}(A)} - \mu(A)\mathbf{1}_{T^{n_1-\max J}(A)} + \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_X \mathbf{1}_{T^{n_1-\max J}(A)}\mathbf{1}_{T^{n_2-\max J}(A)} \, d\mu - \mu(A)^2 \right) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square \end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]

The answer to my question is yes!! [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences." I now give the proof of the answer to my question, in my own words.

First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

Definition. We say that $f_n$ converges uniformly modulo re-ordering to $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$, $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

Proposition. Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$ Then $f_n$ converges uniformly modulo re-ordering to $f$.

Proof. Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$

Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by \begin{align*} f_n(B) &= \mu(A \cap T^{-n}(B)) \\ f(B) &= \mu(A)\mu(B). \end{align*}

Theorem. $f_n$ converges uniformly modulo re-ordering to $f$.

We now prove this, using the above Proposition. For each $N \in \mathbb{N}$, define the probability measure $\mu_N$ on $X^N$ to be the image measure of $\mu$ under the map $x \mapsto (x,T(x),\ldots,T^{N-1}(x))$. For each $n,N \in \mathbb{N}$ with $n \leq N$, let $\pi_{N,\,n} \colon X^N \to X$ be the projection sending $(x_1,\ldots,x_N)$ to $x_n$. Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N \in \mathbb{N}$ with $n \leq N$, \begin{align*} f_n(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mathbf{1}_A \circ \pi_{N,\,N-n} \, d\mu_N \\ f(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mu(A) \, d\mu_N. \end{align*} So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives $$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{\pi_{\max J,\,\max J}^{-1}(B)} \underbrace{\mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_A \circ \pi_{\max J,\,\max J-n}}_{=: \, g_J} \, d\mu_{\max J}. $$ Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

Lemma. Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \| g_J \|_{L^2(\mu_{\max J})} \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

Proof. We have \begin{align*} \| g_J \|_{L^2}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n}) \right\|_{L^2}^2 \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J}} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J}} \mu(A)^2 - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2} - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1} + (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_{X^{\max J}} (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} - \mu(A)^2 \right) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square \end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]

The answer to my question is yes!! [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]

A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): On the mean ergodic theorem for subsequences." I now give the proof of the answer to my question, in my own words.

First some generalities:

Fix a set $S$, a sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.

Definition. We say that $f_n$ converges uniformly modulo re-ordering to $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$, $$ \#\{n \in \mathbb{N} : |f_n(x)-f(x)|\geq\varepsilon \} \leq N. $$

Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.

Proposition. Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \sup_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$ Then $f_n$ converges uniformly modulo re-ordering to $f$.

Proof. Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f_n(x_k)-f(x_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x_{m_k})$ of $(x_k)$ such that, writing $I_k=\{n : f(x_{m_k}) \geq f_n(x_{m_k}) + \varepsilon\}$, we have $\#I_k \to \infty$ as $k \to \infty$. Take a finite $J_k \subset I_k$ for each $k$, such that $\#J_k$ and $\Delta J_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x_{m_k}) - \frac{1}{\#J_k}\!\sum_{n \in J_k} f_n(x_{m_k}) \geq \varepsilon$ for all $k$. $\ \square$

Now a re-statement and proof of the result:

Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by \begin{align*} f_n(B) &= \mu(A \cap T^{-n}(B)) \\ f(B) &= \mu(A)\mu(B). \end{align*}

Theorem. $f_n$ converges uniformly modulo re-ordering to $f$.

We now prove this, using the above Proposition. For each $N \in \mathbb{N}$, define the probability measure $\mu_N$ on $X^{N+1}$ to be the image measure of $\mu$ under the map $x \mapsto (x,T(x),\ldots,T^N(x))$. For each $n,N$ with $0 \leq n \leq N$, let $\pi_{N,\,n} \colon X^{N+1} \to X$ be the projection sending $(x_0,\ldots,x_N)$ to $x_n$. Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$, \begin{align*} f_n(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mathbf{1}_A \circ \pi_{N,\,N-n} \, d\mu_N \\ f(B) &= \int_{\pi_{N,\,N}^{-1}(B)} \mu(A) \, d\mu_N. \end{align*} So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives $$ f(B) - \tfrac{1}{\#J}\!\sum_{n \in J} f_n(B) \ = \ \int_{\pi_{\max J,\,\max J}^{-1}(B)} \underbrace{\mu(A) - \tfrac{1}{\#J}\!\sum_{n \in J} \mathbf{1}_A \circ \pi_{\max J,\,\max J-n}}_{=: \, g_J} \, d\mu_{\max J}. $$ Therefore, to be able to apply the Proposition, it will be sufficient to show the following.

Lemma. Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence: $$ \| g_J \|_{L^2(\mu_{\max J})} \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$

Proof. We have \begin{align*} \| g_J \|_{L^2}^2 &= \left\| \tfrac{1}{\#J}\!\sum_{n \in J} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n}) \right\|_{L^2}^2 \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J+1}} (\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mu(A) - \mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \int_{X^{\max J+1}} \mu(A)^2 - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2} - \mu(A)\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1} + (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \left( \int_{X^{\max J+1}} (\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_1})(\mathbf{1}_A \circ \pi_{\max J, \, \max J-n_2}) \, d\mu_{\max J} - \mu(A)^2 \right) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n_1,n_2 \in J} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &= \left( \tfrac{1}{\#J} \right)^{\!2} \sum_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum_{\substack{\text{distinct} \\ n_1,n_2 \in J}} \big( \mu(A \cap T^{-|n_1-n_2|}(A)) - \mu(A)^2 \big) \\ &\leq \underbrace{\tfrac{1}{\#J}}_{\to 0} \ + \ \underbrace{\sup_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}_{\to 0}. \quad \square \end{align*}

[NB: The fact that $T$ is mixing only entered at the very final step.]