Prove that $\mathbb{Q}$ is dense. Let $a$ and $b$ be the real numbers with $a<b$. Prove that there are integers $m$ and $n\neq 0$ so that
\begin{align*}
a<\frac{m}{n}<b.
\end{align*}
Is my attempt correct?
Proof
Let $x=(b-a)>0$ and $y=1>0$, then there is a positive integer $n$ such that
\begin{align*}
n(b-a)>1\implies nb>na+1 &&\text{By the Archiemedean Property}.
\end{align*}
If $na> 0$ and therefore bounded below, there is a least integer $m$ such that $na<m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m-1<na<m
\end{align*}
Then, we have
\begin{align*}
na<m<na+1<nb\implies na<m<nb
\end{align*}
Note that if $na=0$, the above is still true as its bounded  both above and below, so there is a least integer $m$ such that $m-1<na<m$ by the Well-Ordering Principle. 
If $na < 0$ and therefore bounded above, there is a greatest integer $m$ such that $na>m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m<na<m+1<na+1<nb
\end{align*}
Let $m+1=m'$, we have
\begin{align*}
na<m'<nb
\end{align*}
So for any $x$, the below holds:
\begin{align*}
na<m<nb\implies a<\frac{m}{n}<b &&\text{because $n>0$.}
\end{align*}

I have included some minor corrections in red.
If $na > 0$, there is a least integer $m$ such that $na < m$. This means the below is true
$$m-1 \color{red}{\leq} na < m$$
\begin{align*}
na<m \color{red}{\leq}na+1<nb\implies na<m<nb
\end{align*}
Similar for $na=0$.
If $na < 0$ and therefore bounded above, there is a greatest integer $m$ such that $na \color{red}{\geq}m$ by the Well-Ordering Principle. This means the below is true
\begin{align*}
m\color{red}{\leq}na<m+1\color{red}{\leq}na+1<nb
\end{align*}
Let $m+1=m'$, we have
\begin{align*}
na<m'<nb
\end{align*}
