My question is concerning my proof's validity in the theorem written below. If someone could take a moment and see if there is a flaw in it, I'd appreciate it.
Lemma. There are no natural numbers less than $1$.
Proof. Assume to the contrary that there are natural numbers less than $1$ and let $S$ consist of these numbers. By assumption, $S \neq \varnothing$ and $S \subseteq \mathbb{N}$, so by the well ordering principle, the natural number $m = \min{S}$ exists such that
$$ 0 < m < 1$$
If we multiply the above inequality by $m$, we get
$$0 < m^2 < m < 1$$
We see that $m^2$ is a natural number and it is less than $1$, so it belongs to $S$. But $m^2 < m$, which is a contradiction.
Hence, there are no natural numbers less than $1$.
$\blacksquare$
Theorem. If $m$ and $n$ are integers with $m < n$, then $m \leq n-1$.
Proof. Assume to the contrary that $n-1 < m$. Then
$$n-1 < m < n$$
We observe that the length of this interval $(n-1, n)$ is $1$. If we let
\begin{align} \mathcal{l_1} &= |m - n + 1| \\ \mathcal{l_2} &= |n-m| \end{align}
which are the lengths of the subintervals $(n-1, m]$ and $(m,n)$.
Then because $\mathbb{Z}$ is closed under addition and multiplication, and because $\mathcal{l_1},\mathcal{l_2} > 0$, we see that $\mathcal{l_1}, \mathcal{l_2} \in \mathbb{N}.$
However, $\mathcal{l_1} + \mathcal{l_2} = 1$, so we must have
$$0<\mathcal{l_1} < 1 \thinspace \thinspace \thinspace \thinspace \text{and} \thinspace \thinspace \thinspace \thinspace 0<\mathcal{l_2} < 1$$
which contradict the above lemma that there are no natural numbers less than $1$.
Therefore, we must have $m \leq n-1$ whenever $m<n$.
$\blacksquare$
Corollary. There are no integers between two consecutive integers.
Edit I included a revision of my proof for the lemma, thanks to the commentators below.
Edit 2 I edited the proof to the above lemma to exclude the peano axioms, which is very important to me, since the textbook from which this problem came does not discuss them. The proof is now accessible to anyone beginning elementary number theory.