Theorem If $A$ is any nonempty subset of $\mathbb{Z}^+$, there is some element $m \in A$ such that $m \leq a$, for all $a \in A$ where such $m$ is the minimal element of $A$. (AA: Dummit and Foote)
Proof We use simple induction. Base case: Pick an arbitrary $a_1 \in A$ and construct $A_1 = \{a_1\}$. Clearly the minimal element $m_1 = a_1$. If $A = A_1$ we are done.
If not, choose another $a_2 \in A$ that we didn't already pick and construct $A_2 = \{a_2\} \cup A_1$ and choose $m_2 = $ min$(a_2, m_1)$.
Continue by choosing $a_n$ such that $a_n \in A, a_n \notin A_{n-1}$ and constructing $A_n = \{a_n\} \cup A_{n-1}$, $m_n = $ min$(a_n, m_{n-1})$. We know $A_{n-1}, m_{n-1}$ exist by our inductive hypothesis. If $A_n = A$ then the minimal element is $m_n$ and we are done. $$\tag*{$\blacksquare$}$$
Am I missing something here? I'm aware of many possible correct proofs like in PCP (https://web.archive.org/web/20140825082814/http://crazyproject.wordpress.com:80/2009/12/30/every-nonempty-set-of-positive-integers-has-a-unique-least-element ) but I'd like to know whether my proof is solid.
The only suspicion I can think of is that we have to construct an infinite set $A$ (although countable as $A \subseteq \mathbb{Z}^+$)
This proof is invalid because it assumes $A$ is finite. For example, if $A=\Bbb{Z}^+$, then there is no $A_n$ such that $A=A_n$ because $A_n$ is finite while $A$ is infinite, so this doesn't prove that $A$ has a minimal element.