I have tried to prove the following: "Show that for every real number $x$ there is exactly one integer $N$ such that $N\leq x < N+1$."
What I am wondering is whether or not the proof is correct, assuming that all references I have made to propositions are correct. If something is off, short feedback on what the problem might be will be appreciated.
We shall start by showing existence. We proceed by contradiction. Existence for the case $x=0$ is trivial.
Suppose it is not the case that for every real number $x$ there exists an integer $N$ such that $N\leq x< N+1$. Then for some real number $x$ there does not exist an integer $N$ such that $N\leq x< N+1$. Assuming that $x$ is positive, we know by proposition 5.4.12 and proposition 4.4.1 that there exists an integer such that $N\leq x$. By our assumption about $x$, $N+1\leq x$. Now suppose that $N+k\leq x$ for some natural number $k$. Since $N+k$ is an integer, this implies that $N+k+1\leq x$. We have thus shown by induction that $N+k\leq x$ for all natural numbers $k$. By the Archimedean property, we know that $x\leq M$ for some positive integer $M$. Thus $N+k\leq x \leq M$ for all natural numbers $k$. Letting $k=M+1-N$, we get $M+1\leq x\leq M$, a contradiction. Note that $k$ as chosen above is in fact a natural number. We conclude that for every positive real number $x$ there exists an integer $N$ such that $N\leq x<N+1$.
By proposition 5.4.4 $x$ is positive if and only if $-x$ is negative. Hence, for any negative real number $-x$ we have $-(N+1)< -x\leq -N$. If $-x = -N$, then $-N\leq -x < -N+1$. If not we have $-(N+1)\leq -x < -N$. This concludes the existence part of the proof.
For uniqueness, suppose there exists an integer $M\neq N$ such that $M\leq x<M+1$, and that $N\leq x<N+1$. Suppose first $M<N$. Then $M+1\leq N$, which implies $M+1\leq x$, a contradiction. Similarly, $M>N$ implies $x\leq M$, a contradiction. This concludes the proof.
Thanks in advance to those that take the time to look at this.