This question is about proving the well-ordering principle of the natural numbers using the principle of induction. This other question uses a different induction argument than the one below. I'd like to know if the argument below is correct.
Let $A_n\subset \mathbb{N}$ be a set with cardinality $n$, ie $|A_n|=n$.
Let $f_n:\{1,...,n\}\to A_n$ be some bijection.
For any $k\in\mathbb{N}$ with $k\leq n$, let $f_k:\{1,...,k\}\to A_k$ be the bijection
$$f_k(n)=f_n(n)\ \ n=1,...,k$$
Note that the image of $f_k$ is $A_k$.
Let $P(k)$ be the property that $A_k$ has a smallest element.
Proposition: $P(k)$ is true for $k=1,2,...,n$.
Proof
For $k=1$, $A_1$ has a single element which is also its smallest element.
Assume $P(k)$, ie that $A_k$ has a smallest element $a_k$.
Recall that $A_{k+1}$ is the image of $f_{k+1}$ and that this image contains one more element than $A_k$, namely $f_{k+1}(k+1)=f_n(k+1)$.
Thus,
$$A_{k+1}=A_k\cup\{f_{k+1}(k+1)\}$$
If $f_{n}(k+1) < a_k$ then $f_{n}(k+1)$ is the smallest element of $A_{k+1}$. Otherwise, $a_k$ is.
Thus, $P(k+1)$ is true, ie $A_{k+1}$ has a smallest element.
Thus, by induction, we infer that $P(k)$ is true for $k=1,2,...,n$ and in particular this means that $A_n$ has a smallest element. $\blacksquare$
Finally, since $A_n$ was a generic finite subset of $\mathbb{N}$ then the above reasoning is true for every finite subset of $\mathbb{N}$.
Hence, every finite subset of $\mathbb{N}$ has a smallest element.