[proofplan] We compare the entropy of an arbitrary finite measurable partition of the factor system with the entropy of its pullback partition on the original system. The factor relation implies that every $n$-step name for the pulled-back partition under $T$ is the pullback of the corresponding $n$-step name under $S$, up to $\mu$-null sets. Since $\pi_*\mu=\nu$, the corresponding atoms have the same measures, so their finite-block entropies are equal for every $n$. Taking entropy rates and then the supremum over finite partitions of $Y$ gives the desired inequality. [/proofplan]

[step:Pull back an arbitrary finite partition of the factor system] Let $\mathcal Q=\{Q_1,\dots,Q_m\}$ be a finite measurable partition of $Y$. Define its pullback partition $\pi^{-1}\mathcal Q$ of $X$ by \begin{align*} \pi^{-1}\mathcal Q := \{\pi^{-1}(Q_i):1\le i\le m\}. \end{align*} Since $\pi:X\to Y$ is measurable, each $\pi^{-1}(Q_i)$ belongs to $\mathcal B_X$. Since the sets $Q_i$ are pairwise disjoint and cover $Y$, their preimages are pairwise disjoint and cover $X$. For a finite measurable partition $\mathcal P=\{P_1,\dots,P_r\}$ of a probability space $(Z,\mathcal B,\lambda)$, define its Shannon entropy by \begin{align*} H_\lambda(\mathcal P):=-\sum_{k=1}^{r}\lambda(P_k)\log \lambda(P_k), \end{align*} with the convention $0\log 0=0$. [/step]

[step:Identify the block partition of the pullback with the pullback of the block partition]For each $n\in\mathbb N$, define the $n$-block partition of $\mathcal Q$ under $S$ by \begin{align*} \mathcal Q_n^S := \bigvee_{j=0}^{n-1} S^{-j}\mathcal Q. \end{align*} Define the $n$-block partition of $\pi^{-1}\mathcal Q$ under $T$ by \begin{align*} \mathcal P_n^\top := \bigvee_{j=0}^{n-1} T^{-j}(\pi^{-1}\mathcal Q), \end{align*} where $\mathcal P:=\pi^{-1}\mathcal Q$. For a word $a=(a_0,\dots,a_{n-1})\in\{1,\dots,m\}^n$, define the corresponding atom in $Y$ by \begin{align*} A_a := \bigcap_{j=0}^{n-1} S^{-j}(Q_{a_j}), \end{align*} and define the corresponding atom in $X$ by \begin{align*} B_a := \bigcap_{j=0}^{n-1} T^{-j}\bigl(\pi^{-1}(Q_{a_j})\bigr). \end{align*} The factor relation gives $\pi\circ T=S\circ\pi$ $\mu$-almost everywhere. Let $N\in\mathcal B_X$ be the exceptional set \begin{align*} N:=\{x\in X: \pi(Tx)\ne S(\pi(x))\}. \end{align*} Then $\mu(N)=0$. Because $T$ is measure-preserving, $\mu(T^{-\ell}N)=\mu(N)=0$ for every integer $\ell\ge 0$. For each $j\in\{0,\dots,n-1\}$, define \begin{align*} E_j:=\bigcup_{\ell=0}^{j-1}T^{-\ell}N, \end{align*} with $E_0:=\varnothing$. Then $\mu(E_j)=0$, and induction on $j$ gives \begin{align*} \pi(T^j x)=S^j(\pi(x)) \end{align*} for every $x\in X\setminus E_j$. Let \begin{align*} E_n := \bigcup_{j=0}^{n-1} E_j. \end{align*} Then $\mu(E_n)=0$, and for every $x\in X\setminus E_n$ and every word $a\in\{1,\dots,m\}^n$, \begin{align*} x\in B_a \iff \pi(x)\in A_a. \end{align*} Therefore \begin{align*} \mu\bigl(B_a\triangle \pi^{-1}(A_a)\bigr)=0 \end{align*} for every word $a$.[/step]

[guided]The only point that needs care is that the factor identity may hold only almost everywhere, not pointwise everywhere. We therefore record explicitly where the exceptional sets are, and we also use that $T$ is measure-preserving so that null exceptional sets stay null after pulling them back along iterates of $T$. Fix $n\in\mathbb N$. Let $N\in\mathcal B_X$ denote the exceptional set where the one-step factor identity fails: \begin{align*} N:=\{x\in X: \pi(Tx)\ne S(\pi(x))\}. \end{align*} The factor relation says exactly that $\mu(N)=0$. Since $T$ is measure-preserving, for every integer $\ell\ge 0$ we have \begin{align*} \mu(T^{-\ell}N)=\mu(N)=0. \end{align*} For each time $j\in\{0,\dots,n-1\}$, define \begin{align*} E_j:=\bigcup_{\ell=0}^{j-1}T^{-\ell}N, \end{align*} with $E_0:=\varnothing$. This set removes precisely the points whose orbit hits the one-step exceptional set before time $j$. Because it is a finite union of null sets, $\mu(E_j)=0$. If $x\in X\setminus E_j$, then $T^\ell x\notin N$ for every $\ell\in\{0,\dots,j-1\}$, so the identity $\pi(T^{\ell+1}x)=S(\pi(T^\ell x))$ holds at each of those orbit points. Induction on $\ell$ gives \begin{align*} \pi(T^j x)=S^j(\pi(x)). \end{align*} Since there are only finitely many times in the block, the set \begin{align*} E_n := \bigcup_{j=0}^{n-1}E_j \end{align*} also satisfies $\mu(E_n)=0$. Now fix a word $a=(a_0,\dots,a_{n-1})\in\{1,\dots,m\}^n$. The atom of the $S$-block partition with name $a$ is \begin{align*} A_a := \bigcap_{j=0}^{n-1} S^{-j}(Q_{a_j}), \end{align*} and the atom of the pulled-back $T$-block partition with the same name is \begin{align*} B_a := \bigcap_{j=0}^{n-1}T^{-j}\bigl(\pi^{-1}(Q_{a_j})\bigr). \end{align*} For $x\in X\setminus E_n$, membership in $B_a$ means that $T^j x\in \pi^{-1}(Q_{a_j})$ for every $j$, equivalently \begin{align*} \pi(T^j x)\in Q_{a_j} \end{align*} for every $j\in\{0,\dots,n-1\}$. On $X\setminus E_n$, we may replace $\pi(T^j x)$ by $S^j(\pi(x))$, so this is equivalent to \begin{align*} S^j(\pi(x))\in Q_{a_j} \end{align*} for every $j\in\{0,\dots,n-1\}$. That condition is exactly $\pi(x)\in A_a$, or $x\in \pi^{-1}(A_a)$. Hence $B_a$ and $\pi^{-1}(A_a)$ agree outside the null set $E_n$, so \begin{align*} \mu\bigl(B_a\triangle \pi^{-1}(A_a)\bigr)=0. \end{align*}[/guided]

[step:Use the pushforward identity to match every block atom measure] Since $\pi_*\mu=\nu$, every measurable set $C\in\mathcal B_Y$ satisfies \begin{align*} \mu(\pi^{-1}(C))=\nu(C). \end{align*} Applying this with $C=A_a$, and using the null-set equality from the previous step, gives \begin{align*} \mu(B_a)=\mu(\pi^{-1}(A_a))=\nu(A_a) \end{align*} for every $a\in\{1,\dots,m\}^n$. Thus the atom measures of $\mathcal P_n^\top$ and $\mathcal Q_n^S$ agree, counted with the same word index. Therefore \begin{align*} H_\mu(\mathcal P_n^\top)=H_\nu(\mathcal Q_n^S) \end{align*} for every $n\in\mathbb N$. [/step]

[step:Pass from block entropies to entropy rates] By the definition of entropy with respect to a finite partition, \begin{align*} h_\mu(T,\pi^{-1}\mathcal Q)=\lim_{n\to\infty}\frac{1}{n}H_\mu\left(\bigvee_{j=0}^{n-1}T^{-j}(\pi^{-1}\mathcal Q)\right) \end{align*} and \begin{align*} h_\nu(S,\mathcal Q)=\lim_{n\to\infty}\frac{1}{n}H_\nu\left(\bigvee_{j=0}^{n-1}S^{-j}\mathcal Q\right). \end{align*} The block entropy equality from the previous step gives \begin{align*} h_\mu(T,\pi^{-1}\mathcal Q)=h_\nu(S,\mathcal Q). \end{align*} Since $h_\mu(T)$ is the supremum of $h_\mu(T,\mathcal P)$ over all finite measurable partitions $\mathcal P$ of $X$, and $\pi^{-1}\mathcal Q$ is one such partition, we have \begin{align*} h_\nu(S,\mathcal Q)=h_\mu(T,\pi^{-1}\mathcal Q)\le h_\mu(T). \end{align*} [/step]

[step:Take the supremum over all finite partitions of the factor] The finite partition $\mathcal Q$ of $Y$ was arbitrary. Taking the supremum over all finite measurable partitions $\mathcal Q$ of $Y$ yields \begin{align*} h_\nu(S)=\sup_{\mathcal Q} h_\nu(S,\mathcal Q)\le h_\mu(T). \end{align*} This is exactly the desired entropy monotonicity inequality under factor maps. [/step]