I wanted to cross my t's and dot my i's regarding this below proof.
For any set $S_n=\{x \in \mathbb N \ |\ x \lt n \in \mathbb N \}$, there exists a bijection $f_n: S_n \to \{1,2,\cdots,n-1\}$. Therefore, any such $S_n$ can be called finite.
To prove this, we can proceed with induction on $n$. I was a little concerned about my base case, given how I defined the codomain in the above theorem.
Base Case: $n=1$.
By definition, $S_0 = \emptyset$. Further, no such set of natural numbers satisfies the codomain requirements, so the codomain is also $\emptyset$. Let $f_0 = \emptyset$. Then we have the bijection $f_0: \emptyset \to \emptyset$.
Assume for $n$:
Suppose for any $n \gt 1$, there exists a bijection $f_n: S_n \rightarrow \{1,2,...,n-1\}$.
Prove for $n+1$:
Consider the set $S_{n+1}$. We have $n \in S_{n+1}$ because $n \lt n+1$ and $n \in \mathbb N$. $n$ must be the maximum of $S_{n+1}$ which means all other members of $S_{n+1}$ are less than $n$. But this just means $S_n \subset S_{n+1}$. Therefore, $S_{n+1}=S_n+\{n\}$. Let the bijective function $g$ be defined as $g: \{n\} \to \{n\}$. Now, let $f_{n+1}=f_n \cup g$ and note that $g \cap f_n = \emptyset$. The union of two non-intersecting bijective functions is, itself, a bijective function.