[proofplan]
Fix a finite measurable partition of the target space and pull it back along the measurable map $T$. Intersecting those preimages with the atom $A$ gives a finite measurable partition of $A$. Atomicity forces each piece to have either zero measure or the full measure $\mu(A)$. Finite additivity then rules out two positive pieces, since two full-measure disjoint pieces inside a finite-measure atom would force $\mu(A)\ge 2\mu(A)$.
[/proofplan]
[step:Pull back the target partition to a measurable partition of the atom]
Let $I:=\{1,\dots,m\}$ denote the index set of the partition. For each $i\in I$, define
\begin{align*}
C_i:=A\cap T^{-1}(B_i).
\end{align*}
Since $T:(X,\mathcal A)\to(Y,\mathcal B)$ is measurable and $B_i\in\mathcal B$, the preimage $T^{-1}(B_i)$ belongs to $\mathcal A$. Since $A\in\mathcal A$, closure of $\mathcal A$ under finite intersections gives $C_i\in\mathcal A$ for every $i\in I$.
The sets $(C_i)_{i=1}^m$ are pairwise disjoint because the sets $(B_i)_{i=1}^m$ are pairwise disjoint. Also,
\begin{align*}
\bigcup_{i=1}^m C_i=\bigcup_{i=1}^m \bigl(A\cap T^{-1}(B_i)\bigr)=A\cap T^{-1}\left(\bigcup_{i=1}^m B_i\right)=A\cap T^{-1}(Y)=A.
\end{align*}
Thus $(C_i)_{i=1}^m$ is a finite measurable partition of $A$.
[guided]
We first translate the partition of the target space $Y$ into a partition of the mass sitting inside the atom $A$. Let
\begin{align*}
I:=\{1,\dots,m\}
\end{align*}
be the finite index set, and for each $i\in I$ define the measurable candidate piece
\begin{align*}
C_i:=A\cap T^{-1}(B_i).
\end{align*}
Why is $C_i$ measurable? The partition is measurable, so $B_i\in\mathcal B$. The map $T:(X,\mathcal A)\to(Y,\mathcal B)$ is measurable, so $T^{-1}(B_i)\in\mathcal A$. Since $A\in\mathcal A$ and $\mathcal A$ is closed under finite intersections, we get $C_i\in\mathcal A$.
Next we check that these pieces really form a partition of $A$. If $i,j\in I$ with $i\ne j$, then $B_i\cap B_j=\varnothing$ because $(B_i)_{i=1}^m$ is a partition of $Y$. Therefore
\begin{align*}
C_i\cap C_j=A\cap T^{-1}(B_i)\cap A\cap T^{-1}(B_j)=A\cap T^{-1}(B_i\cap B_j)=A\cap T^{-1}(\varnothing)=\varnothing.
\end{align*}
Finally, the pieces cover $A$ because the sets $B_i$ cover $Y$:
\begin{align*}
\bigcup_{i=1}^m C_i=\bigcup_{i=1}^m \bigl(A\cap T^{-1}(B_i)\bigr)=A\cap T^{-1}\left(\bigcup_{i=1}^m B_i\right)=A\cap T^{-1}(Y)=A.
\end{align*}
Thus the pulled-back pieces $(C_i)_{i=1}^m$ are measurable, pairwise disjoint, and have union $A$.
[/guided]
[/step]
[step:Apply atomicity to each pulled-back piece]
For every $i\in I$, the set $C_i$ is a measurable subset of $A$. Since $A$ is an atom of $\mu$, the defining property of an atom gives
\begin{align*}
\mu(C_i)\in\{0,\mu(A)\}
\end{align*}
for every $i\in I$.
[/step]
[step:Use finite additivity to rule out two positive pieces]
Assume, toward a contradiction, that there exist distinct indices $i,j\in I$ such that
\begin{align*}
\mu(C_i)>0 \quad \text{and} \quad \mu(C_j)>0.
\end{align*}
By the previous step, positive measure forces
\begin{align*}
\mu(C_i)=\mu(A) \quad \text{and} \quad \mu(C_j)=\mu(A).
\end{align*}
Since the sets $(C_k)_{k=1}^m$ are pairwise disjoint and have union $A$, finite additivity of the measure $\mu$ gives
\begin{align*}
\mu(A)=\sum_{k=1}^m \mu(C_k).
\end{align*}
All terms in the sum are nonnegative, so
\begin{align*}
\mu(A)=\sum_{k=1}^m \mu(C_k)\ge \mu(C_i)+\mu(C_j)=2\mu(A).
\end{align*}
Because $0<\mu(A)<\infty$, the inequality $\mu(A)\ge 2\mu(A)$ is impossible. Hence no two distinct indices can satisfy $\mu(C_i)>0$.
Substituting back $C_i=A\cap T^{-1}(B_i)$, at most one index $i\in\{1,\dots,m\}$ satisfies
\begin{align*}
\mu(A\cap T^{-1}(B_i))>0.
\end{align*}
This proves the theorem.
[/step]
