Let $\mathcal{A}$ be the system of non-empty sets. Then $\prod{\mathcal{A}} = \prod_{B \in \mathcal{A}}{B}$ is non-empty set.
note: $\prod{\mathcal{A}}$ consists of functions $f:\mathcal{A} \rightarrow \cup\mathcal{A}$ satisfying $f(B) \in B, \forall B \in \mathcal{A}$.
My question is about that note, I probably don't understand it, because let $A = \{1\}, B = \{2\}$, then $\prod{\mathcal{A}} = \prod_{B \in \mathcal{A}}{B} = A\times B = \{[1, 2]\}$, so the product of $A$ and $B$ is only one element set containing $[1, 2]$ and definitely not $\{[A, 1], [B, 2]\}$.
The product of $\{\{1\},\{2\}\}$ has exactly one element. The function $f$ such that $f(\{1\})=1$ and $f(\{2\})=2$.
This is a small problem we have when we think about Cartesian products and choice functions and tuples as literally the same objects. They are not the same objects, but there is a natural way to translate one object to another.
Ordered pairs are very specific and well-defined sets coding the notion of an ordered pair into sets;
Tuples are functions from an index set into some set;
Choice functions are functions taking a set and returning an element of that set, for every set in their domain.
If we have an indexed family, especially of two sets, then it is easy to confuse all three definitions of ordered pairs, $2$-tuples, and choice functions. They are not literally the same thing, but we can easily translate between them, so we often use the same notation for all the relevant sets and call it "their product".