I'm wondering if the next argument is sound or maybe need some adjustments;
Proposition (Interspersing of integers by rationals): Let $x\in \mathbb{Q}$. Then there exists an integer such that $n\le x< n+1$. In fact this integer is unique. In particular, there exists a natural number $N$ such that $N>x$ (i.e. there is no such thing as a rational number which is larger than all the natural numbers).
Proof: By the trichotomy property of the rational numbers, either $x$ is positive, negative or zero.
If $x=0$, it's trivial since $\,0 \le x<1$.
Now if $x$ is positive number, by definition $x= \frac{p}{q}$ where $p,q\in \mathbb{N}- \left\{0 \right\}$. Then using the Euclidean algorithm we have that:
$$p = mq+r, \text{ where } \:\: 0 \le r < q $$
We shall show that the natural number $m$ has the desired property $\,m \le x < m+1$. It is sufficient to show that $\,\,0 \le x-m<1$.
$x -m= \frac{p}{q}-m =\frac{mq+r}{q}-{m} = \frac{r}{q}$. Since by the Euclidean algorithm we know that $\,0 \le r < q$ then $\,0 \le \frac{r}{q} <1$ as desired.
Now suppose there is a $n \not= m $ such that $p = nq+r$. So, $\,mq+r = nq+r$ and we conclude that $m=n$ which show the uniqueness of $m$.
If $x$ is a negative rational number, we have that $x = \frac{-a\,}{\,\,\,b\,}$ where $a,b\in \mathbb{N}- \left\{0\right\}$, so $-x = \frac{a}{b}$ and we may use the same argument that in the positive case.
To conclude we need to show that the integer is unique and is follows of the above argument.
I feel that the argument is a bit flawed, I would appreciate any suggestion. Thanks in advance
If you want that $m$ is a natural number, then we should have $p\ge q$. If $p<q$, and both positive, then anyway $0\le p/q<1$. Otherwise the proof should work like what you did. It is enough to consider the case that $x$ is positive, by the way.