This is exercise 4 from chapter I 3.12 from Tom Apostol's "Calculus" (2nd edition):
If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies the inequalities $n \le x \lt n+1$. This $n$ is called the greatest integer in $x$ and is denoted by $\lfloor x \rfloor$.
Up to this point in the book induction, nor well-ordering principle isn't introduced. All that's available is some basic theorems deduced from 10 axioms of real numbers. For example I can't claim that for every real number $x$ there is $n$ satisfying $x \le n \lt x + 1$ (which would be enough for me to make the proof).
My aim at the proof was:
Let S be the set of all positive integers $n$ which satisfy $n \lt x$.
From Theorem I.29 (For every real $x$ there exists a positive integer $n$ such that $n \gt x$) S is not empty.
Notice that $y < x$ for every $y \in S$, so $x$ is an upper bound of $S$.
From Axiom 10 (Every upper-bounded set of real numbers has a supremum) $S$ has supremum $L$ and $L \le x$.
I don't know how to prove that this supremum is an integer.