elementary number theory - Proving that there does not exist an infinite descending sequence of naturals using minimal counterexample

This might be a long-winded way of proving something so obvious, but I want to have it checked if it holds up.

Claim: There does not exist a sequence $\{{a}_{n \in \mathbb{N}}\}$ of natural numbers such that $a_{n+1} < a_{n}$ for every $n$.

Proof: This amounts to showing that the set $M = \{ m \in \mathbb{N} : m \leq a_{0}\}$, where $a_{0}$ is the first member from any sequence in infinite descent, is finite. Suppose for the sake of contradiction that there exists an $a_{0}$ such that the set $M$ is infinite. We suppose further that $a_{0}$ is minimal. Clearly $a_{0} \neq 0$, for if it were, the set $M = \{ m \in \mathbb{N} : m \leq 0\}$ = $\{0\}$ is finite.

Consider the set $N = M - \{n \in \mathbb{N}: a_{1} < n \leq a_{0}\}$, which is just the set $M$ with the finitely many naturals deleted. From elementrary set theory, $N$ must be infinite.

But $N = \{ n \in \mathbb{N} : n \leq a_{1}\}$, and we can suppose that $a_{1}$ is the first term in a natural number sequence in infinite descent. Thus we have found another infinite set with the desired property, but with $a_{1} < a_{0}$. This contradicts the minimality of $a_{0}$.

Answer

This amounts to showing that the set $M = \{ m \in \mathbb{N} : m \leq a_{0}\}$, where $a_{0}$ is the first member from any sequence in infinite descent, is finite.
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\def\nn{\mathbb{N}}
$

This first line of your 'proof' is invalid. You are using merely your intuition to claim that the non-existence of an infinite descending chain of natural numbers follows from the finiteness of some set that you specified. This is circular in this case; proving the equivalence amounts to proving something of roughly the same strength as the original desired theorem.

Instead what you actually need is:

Let $S = \{ n : n \in \nn \land \text{there is a strictly decreasing sequence from $\nn$ starting with $n$ } \}$.

If $S$ is non-empty then let $m = \min(S)$ and (use your other ideas) to prove that there is a strictly decreasing sequence from $\nn$ starting with something less than $m$, which contradicts the definition of $m$.

Therefore $S$ is empty and you are done.