The above proof is from an introductory paper on ordinals. I'm confused by the proof since it seems to assume that every subset of $\mathbb{N}$ has a smallest element, which is something the paper doesn't prove. I went online looking for proofs that $\mathbb{N}$ is well-ordered, but all the ones I've found use induction in one way or another, so I wonder how one could prove that $\mathbb{N}$ is well-ordered without appealing to induction.
In case it is relevant, the paper defines $\mathbb{N}$ by means of the Axiom of Infinity:
You can't really do that.
For a partial order $(P,<)$ we say that $P$ has "capacity for induction"1 if whenever $A\subseteq P$ is such that all the minimal elements are in $A$, and whenever $a\in P$ is such that $\{x\in A\mid x<a\}\subseteq A$, then $a\in A$, then $A=P$.
Theorem. $(P,<)$ has capacity for induction if and only if for every non-empty $A\subseteq P$, there is a minimal element in $A$ (relative to the restricted order).
In other words, capacity for induction is equivalent to being well-founded.
Corollary. If a $P$ is a linear order, then it has the capacity for induction if and only if it is well-ordered.
This means that we can't prove that $\Bbb N$ has the capacity for induction without using the fact that it is well-ordered. Luckily we are not working with the Peano axioms, we are working in $\sf ZF$, where we can prove the fact that our set $\Bbb N$ is in fact well-ordered without appealing to induction principles internal to $\Bbb N$.
Finally, as a small side note, the paper you're reading should state that $\Bbb N$ is not just "a set" with these properties, but in fact the smallest such set.