Let me suggest the most general version of this phenomenon. Consider a family $A$ of sets indexed by index set $I$. Agreeing to denote the set of all partitions of a given set $M$ by $\mathscr{Part}(M)$, let us consider a family $\mathscr{P} \in \displaystyle\prod_{i \in I}\mathscr{Part}(A_i)$ of partitions, which means that the component at each index $i \in I$ is a partition $\mathscr{P}_i \in \mathscr{Part}(A_i)$ on $A_i$.
We proceed to define the following set:
$$\bigotimes_{i \in I}\mathscr{P}_i \colon=\left\{\prod_{i \in I}X_i\right\}_{X \in \displaystyle\prod_{i \in I}\mathscr{P}_i}$$
and seek to prove that:
$$\displaystyle\bigotimes_{i \in I}\mathscr{P}_i \in \mathscr{Part}\left(\displaystyle\prod_{i \in I}A_i\right).$$
To prove this assertion, let us set $\mathscr{Q}\colon=\displaystyle\bigotimes_{i \in I}\mathscr{P}_i$ respectively $B\colon=\displaystyle\prod_{i \in I}A_i$. There are three statements we need to establish separately:
- Any $U \in \mathscr{Q}$ is a nonempty subset $\varnothing \neq U \subseteq B$: by definition, we have the expression $U=\displaystyle\prod_{i \in I}X_i$ for a certain family $X \in \displaystyle\prod_{i \in I}\mathscr{P}_i$. Since $\mathscr{P}_i$ is a partition of $A_i$ we infer $X_i \subseteq A_i$ for every $i \in I$ and thus $U \subseteq B$ (by virtue of the monotony of the cartesian product operation). Likewise, since $X_i \in \mathscr{P}_i$ for each $i \in I$ it follows that $X_i \neq \varnothing$ and as cartesian products of nonempty sets remain nonempty we gather that $U \neq \varnothing$.
- For any $U, V \in \mathscr{Q}$ we have $U \neq V \Rightarrow U\cap V=\varnothing$. Once again, let us consider expressions $U=\displaystyle\prod_{i \in I}X_i$ and $V=\displaystyle\prod_{i \in I}Y_i$, afforded by families $X, Y \in \displaystyle\prod_{i \in I}\mathscr{P}_i$. As $U \neq V$, there must exist an index $i \in I$ such that $X_i \neq Y_i$. However, since $X_i, Y_i$ are distinct members of the partition $\mathscr{P}_i$, this entails $X_i \cap Y_i=\varnothing$ and hence:
$$U \cap V=\left(\prod_{i \in I}X_i\right) \cap \left(\prod_{i \in I}Y_i\right)=\prod_{i \in I}\left(X_i \cap Y_i\right)=\varnothing,$$
bearing in mind that the cartesian product of a family of sets at least one component of which is empty is itself (the product) empty.
- Finally, $B=\bigcup\mathscr{Q}$. Here we shall rely on the following fundamental:
Proposition. Let $J$ be a family of sets indexed by set $I$ and let furthermore $X$ be another family of sets, indexed by the disjoint union $\displaystyle\bigsqcup_{i \in I}J_i\colon=\displaystyle\bigcup_{i \in I}\left(\{i\} \times J_i\right)$. We then have the relation:
$$\prod_{i \in I}\bigcup_{j \in J_i}X_{ij}=\bigcup_{k \in \prod_{i \in I}J_i}\prod_{i \in i}X_{ik_i}.$$
We apply the general proposition above to the instance of family $\mathscr{P}$ of partitions playing the role of $J$ and family $M$ -- given by $M_{iX}=X$, for any $i \in I$ and $X \in \mathscr{P}_i$ -- playing the role of $X$, in order to infer that:
$$\begin{align}
B&=\prod_{i \in I}A_i\\
&=\prod_{i \in I}\bigcup\mathscr{P}_i\\
&=\prod_{i \in I}\bigcup_{X \in \mathscr{P}_i}X\\
&=\bigcup_{Y \in \prod_{i \in I}\mathscr{P}_i}\prod_{i \in Y}M_{iY_i}\\
&=\bigcup_{Y \in \prod_{i \in I}\mathscr{P}_i}\prod_{i \in I}Y_i\\
&=\bigcup_{U \in \mathscr{Q}}U\\
&=\bigcup\mathscr{Q}.
\end{align}$$
This concludes the argumentation and signifies the existence of a natural map:
$$\begin{align}
\prod_{i \in I}\mathscr{Part}(A_i) &\to \mathscr{Part}\left(\prod_{i \in I}A_i\right)\\
\mathscr{P} &\mapsto \bigotimes_{i \in I}\mathscr{P}_i.
\end{align}$$
