Proof of greatest integer theorem: floor function is well-defined I have to prove that

$$\forall x \in \mathbb{R},\exists !\,n \in \mathbb{Z} \text{ s.t. }n \leq x < n+1\;.$$

where $\exists !\,n $ means there exists a unique (exactly one) $n$.
I'm done with proving that there are at least one integers for the solution.
I couldn't prove the "uniqueness" of the solution, and so I looked up the internet, and here's what I found:

Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x < n+1$ and $m \leq x < m+1$.
Since $n \leq x \text{ and } -(m+1) < -x$, by adding both, $n + (-m-1) < (x-x) = 0$. And (some steps here) likewise, $(x-x) < n+m+1$.

Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)?
Or should I also prove those addition of inequalities?
Thank you :D

The usual proof in the context of real analysis goes like this:
Let $A= \{ n \in \mathbb Z : n \le x \}$. Then $A$ is not empty. Indeed, there is $n\in \mathbb N$ such that $n>-x$, because $\mathbb N$ is unbounded. But then $-n\in A$.
Let $\alpha=\sup A$. Then there is $n\in A$ such that $\alpha-1<n\le\alpha$. But then $\alpha<n+1\le\alpha+1\le x+1$ and so $n\le x$ and $n+1\notin A$, that is, $n\le x < n+1$.
If $m\in A$ then $m\le n$ because $m>n$ implies $m\ge n+1>x$. If $m<n$ then $m+1\le n\le x$ and $m$ cannot be a solution. Hence the solution is unique.
