= Solution
Suppose first that $|A_n|=1$ for every $n>N$. Then an element of $B$ is determined by its first $N$ coordinates, so $B$ is in bijection with
$$
A_1\times\cdots\times A_N.
$$
A finite <Cartesian product> of countable sets is countable by repeated application of the diagonal enumeration in part (b)(i). Hence $B$ is countable.
Conversely, suppose infinitely many factors contain at least two elements. Choose increasing indices $n_1<n_2<\cdots$ and distinct elements
$$
a_j^0,a_j^1\in A_{n_j}.
$$
Fix one element in every remaining factor. Each <binary sequence> $\varepsilon=(\varepsilon_1,\varepsilon_2,\ldots)$ then defines an element of $B$ by placing $a_j^{\varepsilon_j}$ in coordinate $n_j$ and the fixed element elsewhere. This map is injective.
The set $\{0,1\}^{\mathbb N}$ is uncountable by <Cantor's diagonal argument>: from any proposed list of binary sequences, form a new sequence whose $j$th digit differs from the $j$th digit of the $j$th listed sequence. It is absent from the list. Thus $B$ contains an uncountable subset and cannot be countable.
Therefore
$$
\boxed{
B\text{ is countable }\Longleftrightarrow
\exists N\ \forall n>N,\ |A_n|=1}.
$$
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