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Write the sets as $A_1,A_2,\dots$, and list each nonempty countable set as $A_i=\{a_{i1},a_{i2},\dots \}$. Enumerate the pairs $(i,j)\in {\mathbb{N}}^2$ along successive diagonals, as in the Cantor pairing function, and output $a_{ij}$ when it has not appeared before. Every member of ${\bigcup }_i{A}_i$ eventually appears, proving that a countable union of countable sets is countable.

$A$ is uncountable. For each binary sequence $({\epsilon }_j)$, defineThis is a nondecreasing function, and different binary sequences give different functions, so the uncountable set of binary sequences injects into $A$.

$B$ is countable. Every nonincreasing sequence of positive integers can decrease only finitely many times and is therefore eventually constant. Such a sequence is specified by a finite sequence of positive integers, and the set of all finite integer sequences is a countable union of countable sets. Hence $B$ is countable.

True. If $\mathrm{TC}(\{x\})$ is countable, then each of its members is a subset of this transitive set and is therefore countable.

Conversely, if $x$ is reasonable, then $x$ itself and every set occurring at every finite membership depth below it are countable. Starting with the countable set $\{x\}$, each next level is a countable union of countable sets, hence countable. Their countable union is $\mathrm{TC}(\{x\})$, so the transitive closure is countable.

A set is countable set if it is finite or admits an enumeration by natural numbers. If $A_n$ are countable, choose enumerations $a_{n,m}$. Since ${\mathbb{N}}^2$ is countable by diagonal enumeration, the array $(n,m)\mapsto a_{n,m}$ enumerates ${\bigcup }_n{A}_n$ after repetitions are removed. This proves that a countable union of countable sets is countable.

The integers are countable, so $\mathbb{Z}\times {\mathbb{N}}_{>0}$ is countable. Mapping $(p,q)$ to $p/q$ surjects onto $\mathbb{Q}$, hence $\mathbb{Q}$ is countable. The same diagonal argument shows that $A\times B$ is countable whenever $A,B$ are.

Finally, if the reals in $(0,1)$ had decimal expansions $x_1,x_2,\dots$, choose a decimal whose $n$th digit differs from the $n$th digit of $x_n$, avoiding digits $0$ and $9$ to remove expansion ambiguity. This number differs from every listed number. The Cantor diagonal argument proves that $\mathbb{R}$ is uncountable.