Let $(A_i)_{i \ge 0}$ be a countable family of sets and $A = \cup A_i$.
For each $i \ge 0$ let there be given $([a_i]_k)_{\,k \ge 1}$, a bijective enumeration of the set $A_i$.
We will be using the following result (see it also used over here).
Proposition 1: A set $A$ is countably infinite if there exist a family of subsets of $A$,
$(A_n)_{\, n \ge 0}$ satisfying
$\quad |A_n| = n$
$\quad A_k \subset A_{k+1} \; \text{ for } k \ge 0$
$\quad A = \bigcup_k A_k$
Moreover, the chain $A_0 \subset A_1 \subset A_2 \dots$ defines an explicit bijective enumeration of the set $A$.
Proof
Given such a chain of finite subsets, for every $k \ge 1$ there exist one and only one element $a_k \in A_k$ such that $a_k \notin A_{k-1}$. One can also easily show that $(a_k)_{\,k \ge 1}$ is a bijective enumeration of $A$.
$\quad \blacksquare$
For each $i \ge 0$ and $m \ge 1$ we define
$\quad A_{(i,m)} = \{ x \in A_i \,| \, x \in \{([a_i]_1,([a_i]_2,\dots,([a_i]_m\}\}$
We define
$$\quad R_m = \bigcup_{i=1}^m A_{(i,m)}$$
If $B$ is any subset of $A$ and if $R_m$ is not included in $B$ we can apply an operator $\mathcal R_m$ to $B$ as follows:
$\quad B \mapsto B \cup \{\hat r\}$
where $\hat r$ is 'cranked-out' as follows:
$\text{1. }$ Find the smallest integer $k$ with $1 \le k \le m$ such that $A_{(k,m)}$ is not contained in $B$.
$\text{2. }$ Find the smallest integer $j$ with $1 \le m \le m$ such that $[a_k]_j \notin R_m$; set $\hat r = [a_k]_j$.
If $C$ is any proper subset of $A$ we define $\mathcal M$ on $C$ as follows:
$\quad C \mapsto m \; \text{ where } m \text{ is the smallest positive integer such that } R_m \text{ is not included in } C$
All the constructions are now available to define the recursion. With $D_0 = \emptyset$ and given $D_k \subsetneq A$ with $k \ge 0$, define
$$ D_{k+1} = \mathcal R_{\mathcal M (D_k)}(D_k)$$
It is easy to see that this chain satisfies the conditions stated in proposition 1.