[proofplan]
We compare the entropy of an arbitrary finite measurable partition of the factor space $Y$ with the entropy of its pullback partition on $X$. The factor relation identifies each finite dynamical join on $Y$ with the corresponding pullback join on $X$, up to null sets, and the condition $\pi_\#\mu=\nu$ makes the atom measures equal. Therefore the entropy generated by any partition of $Y$ is realised by a finite partition of $X$, and taking suprema gives the desired inequality.
[/proofplan]
[step:Pull back an arbitrary finite partition of the factor space]
Let $\mathcal Q=\{Q_1,\dots,Q_m\}$ be a finite $\mathcal C$-measurable partition of $Y$. Define its pullback partition $\pi^{-1}\mathcal Q$ of $X$ by
\begin{align*}
\pi^{-1}\mathcal Q := \{\pi^{-1}(Q_j):1\le j\le m\}.
\end{align*}
Each set $\pi^{-1}(Q_j)$ belongs to $\mathcal B$ because $\pi$ is $(\mathcal B,\mathcal C)$-measurable. Since the sets $Q_j$ are pairwise disjoint and cover $Y$, their preimages are pairwise disjoint and cover $X$. Hence $\pi^{-1}\mathcal Q$ is a finite $\mathcal B$-measurable partition of $X$.
For a finite measurable partition $\mathcal P=\{P_1,\dots,P_r\}$ of a probability space $(Z,\mathcal A,\lambda)$, define its Shannon entropy by
\begin{align*}
H_\lambda(\mathcal P):=-\sum_{i=1}^r \lambda(P_i)\log \lambda(P_i),
\end{align*}
with the convention $0\log 0=0$. Because $\pi_\#\mu=\nu$, for every $1\le j\le m$ we have
\begin{align*}
\mu(\pi^{-1}(Q_j))=\nu(Q_j).
\end{align*}
Therefore
\begin{align*}
H_\mu(\pi^{-1}\mathcal Q)=H_\nu(\mathcal Q).
\end{align*}
[/step]
[step:Identify the dynamical joins under the factor relation]
For each $n\in\mathbb N$, define the $n$-step dynamical join of $\mathcal Q$ under $S$ by
\begin{align*}
\mathcal Q_n^S:=\bigvee_{k=0}^{n-1}S^{-k}\mathcal Q,
\end{align*}
and define the $n$-step dynamical join of $\pi^{-1}\mathcal Q$ under $T$ by
\begin{align*}
(\pi^{-1}\mathcal Q)_n^\top:=\bigvee_{k=0}^{n-1}T^{-k}(\pi^{-1}\mathcal Q).
\end{align*}
An atom of $\mathcal Q_n^S$ has the form
\begin{align*}
A_{j_0,\dots,j_{n-1}}:=\bigcap_{k=0}^{n-1}S^{-k}(Q_{j_k}),
\end{align*}
where each $j_k\in\{1,\dots,m\}$. Its pullback is
\begin{align*}
\pi^{-1}(A_{j_0,\dots,j_{n-1}})=\bigcap_{k=0}^{n-1}\pi^{-1}(S^{-k}(Q_{j_k})).
\end{align*}
Since $\pi\circ T=S\circ\pi$ $\mu$-almost everywhere and both systems are measure-preserving, for every $k\in\{0,\dots,n-1\}$ and every $Q\in\mathcal C$,
\begin{align*}
\pi^{-1}(S^{-k}Q)=T^{-k}(\pi^{-1}Q)
\end{align*}
up to a $\mu$-null set. Therefore the atoms of $\pi^{-1}(\mathcal Q_n^S)$ and the atoms of $(\pi^{-1}\mathcal Q)_n^\top$ agree up to $\mu$-null sets. Entropy of a finite partition depends only on the measures of its atoms, so
\begin{align*}
H_\mu((\pi^{-1}\mathcal Q)_n^\top)=H_\mu(\pi^{-1}(\mathcal Q_n^S)).
\end{align*}
Using $\pi_\#\mu=\nu$ again gives
\begin{align*}
H_\mu(\pi^{-1}(\mathcal Q_n^S))=H_\nu(\mathcal Q_n^S).
\end{align*}
Thus, for every $n\in\mathbb N$,
\begin{align*}
H_\mu((\pi^{-1}\mathcal Q)_n^\top)=H_\nu(\mathcal Q_n^S).
\end{align*}
[guided]
We want to compare the entropy growth of $\mathcal Q$ under $S$ with the entropy growth of $\pi^{-1}\mathcal Q$ under $T$. For each $n\in\mathbb N$, the partition that records the first $n$ iterates of a point in $Y$ is
\begin{align*}
\mathcal Q_n^S:=\bigvee_{k=0}^{n-1}S^{-k}\mathcal Q.
\end{align*}
The corresponding partition on $X$ is
\begin{align*}
(\pi^{-1}\mathcal Q)_n^\top:=\bigvee_{k=0}^{n-1}T^{-k}(\pi^{-1}\mathcal Q).
\end{align*}
The key point is that the factor map transports the $T$-itinerary of $x\in X$ into the $S$-itinerary of $\pi(x)\in Y$. Let
\begin{align*}
A_{j_0,\dots,j_{n-1}}:=\bigcap_{k=0}^{n-1}S^{-k}(Q_{j_k})
\end{align*}
be an atom of $\mathcal Q_n^S$, where each $j_k\in\{1,\dots,m\}$. Its pullback consists of those $x\in X$ such that $\pi(x)$ lies in $Q_{j_0}$, $S(\pi(x))$ lies in $Q_{j_1}$, and so on through time $n-1$. Formally,
\begin{align*}
\pi^{-1}(A_{j_0,\dots,j_{n-1}})=\bigcap_{k=0}^{n-1}\pi^{-1}(S^{-k}(Q_{j_k})).
\end{align*}
Because $\pi\circ T=S\circ\pi$ $\mu$-almost everywhere, iterating gives $\pi\circ T^k=S^k\circ\pi$ $\mu$-almost everywhere for each $k\in\{0,\dots,n-1\}$. Since $Q_{j_k}\in\mathcal C$, this implies
\begin{align*}
\pi^{-1}(S^{-k}(Q_{j_k}))=T^{-k}(\pi^{-1}(Q_{j_k}))
\end{align*}
up to a $\mu$-null set. Intersecting over the finitely many indices $k$ preserves equality up to a $\mu$-null set, so the pullback atom $\pi^{-1}(A_{j_0,\dots,j_{n-1}})$ agrees up to a $\mu$-null set with the corresponding atom of $(\pi^{-1}\mathcal Q)_n^\top$.
Entropy is computed only from the measures of atoms. Changing finitely many atoms by null sets does not change those measures. Therefore
\begin{align*}
H_\mu((\pi^{-1}\mathcal Q)_n^\top)=H_\mu(\pi^{-1}(\mathcal Q_n^S)).
\end{align*}
Finally, the pushforward identity $\pi_\#\mu=\nu$ says that every measurable set $A\in\mathcal C$ satisfies $\mu(\pi^{-1}(A))=\nu(A)$. Applying this to each atom of $\mathcal Q_n^S$ gives
\begin{align*}
H_\mu(\pi^{-1}(\mathcal Q_n^S))=H_\nu(\mathcal Q_n^S).
\end{align*}
Combining the two equalities yields
\begin{align*}
H_\mu((\pi^{-1}\mathcal Q)_n^\top)=H_\nu(\mathcal Q_n^S)
\end{align*}
for every $n\in\mathbb N$.
[/guided]
[/step]
[step:Pass from finite-time entropy to partition entropy]
The entropy of $S$ relative to $\mathcal Q$ is
\begin{align*}
h_\nu(S,\mathcal Q):=\lim_{n\to\infty}\frac{1}{n}H_\nu(\mathcal Q_n^S),
\end{align*}
and the entropy of $T$ relative to $\pi^{-1}\mathcal Q$ is
\begin{align*}
h_\mu(T,\pi^{-1}\mathcal Q):=\lim_{n\to\infty}\frac{1}{n}H_\mu((\pi^{-1}\mathcal Q)_n^\top).
\end{align*}
The finite-time equality from the previous step gives equality of the two sequences term by term. Therefore
\begin{align*}
h_\nu(S,\mathcal Q)=h_\mu(T,\pi^{-1}\mathcal Q).
\end{align*}
[/step]
[step:Take the supremum over finite partitions]
By definition,
\begin{align*}
h_\nu(S):=\sup_{\mathcal Q} h_\nu(S,\mathcal Q),
\end{align*}
where the supremum is taken over all finite $\mathcal C$-measurable partitions $\mathcal Q$ of $Y$. Similarly,
\begin{align*}
h_\mu(T):=\sup_{\mathcal P} h_\mu(T,\mathcal P),
\end{align*}
where the supremum is taken over all finite $\mathcal B$-measurable partitions $\mathcal P$ of $X$.
For each finite $\mathcal C$-measurable partition $\mathcal Q$ of $Y$, the pullback $\pi^{-1}\mathcal Q$ is one of the finite $\mathcal B$-measurable partitions over which the supremum defining $h_\mu(T)$ is taken. Hence
\begin{align*}
h_\nu(S,\mathcal Q)=h_\mu(T,\pi^{-1}\mathcal Q)\le h_\mu(T).
\end{align*}
Taking the supremum over all finite $\mathcal C$-measurable partitions $\mathcal Q$ gives
\begin{align*}
h_\nu(S)\le h_\mu(T).
\end{align*}
This is the desired monotonicity of Kolmogorov-Sinai entropy under factor maps.
[/step]
