You can say that theorems in mathematics are like 'strategic acquisitions'. But even if the OP is lacking the theoretical underpinnings, we can still rigorously describe a bijection between $\mathbb N$ and $\mathbb Q$, and describe a tactical context.
We can 'acquire' a countably infinite set by 'building it up', one element at a time, from finite sets; here is the formal statement
Proposition 1: A set $A$ is countably infinite if there exist a family of subsets of $A$,
$(A_n)_{\, n \ge 0}$ satisfying
Moreover, the chain $A_0 \subset A_1 \subset A_2 \dots$ defines an explicit bijective enumeration of the set $A$.
$\quad |A_n| = n$
$\quad A_k \subset A_{k+1} \; \text{ for } k \ge 0$
$\quad A = \bigcup_k A_k$
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 the OP's problem, we want an (recursive) algorithm to create these subsets. Every finite set of $\Bbb Q$ is bounded, so tactically you want to 'crank-out' sets contained in $[-m, +m]$ and at some point widen the interval and simultaneously get better at 'adding in the still missing numbers'.
We are going to write the proof in the following section, using definitions to make the argument easier to follow. The detail seems appropriate since the OP's question has the tag proof-writing.
Definition: If $R$ is a nonempty finite subset $\Bbb Q$ then let $m$ be the smallest positive integer $m$ such that $R \subset [-m, +m]$. We call this interval the integer interval container for $R$.
For any $m \gt 0$ define the finite sets
$\quad R_m = \{ \frac{n}{d} \, | \, n \in \Bbb Z \land [1 \le d \le m] \land [-m \le \frac{n}{d} \le m]\}$
It is easy to see that the integer interval container for $R_m$ is $[-m, +m]$.
Let $R$ be any finite subset of $\mathbb Q$ with integer interval container $[-m, +m]$. If $R$ doesn't completely contain $R_m$, let $\mathcal{A}(R) = t$ where $t$ is the smallest number in $R_m$ such that $t \notin R$.
We are now ready to recursively construct our chain of subsets (see proposition 1), and $m$ will be associated with the integer interval container construct.
With $A_1 = \{0\}$ and given $A_k$ with $k \ge 0$, define
$$
A_{k+1} = \left\{\begin{array}{lr}
A_k \, \cup \, \{m+1\}, & \text{when } R_m \subset A_k \\
A_k \, \cup \, \mathcal{A}(A_k) , & \text{when } \mathcal{A} \text{ can be applied}
\end{array}\right\}
$$
Setting $A_0 = \emptyset$, this recursion, produces just what we need to apply proposition 1. Here are the first few terms
$A_1 = \{0\} \quad m = 1$
$A_2 = \{-1, 0\}$
$A_3 = \{-1, 0,1\} \quad \text{contains } R_1$
$A_4 = \{-1, 0,1,2\} \quad m = 2$
$A_5 = \{-2, -1, 0,1,2\}$
$A_6 = \{-2,-\frac{3}{2}, -1, 0,1,2\}$
and just keep turning the 'recursive crank'