Prove that every real number belongs to the interval of integers Here is a problem:

Proposition. Prove that for every real number $x$, there exists a unique integer $N$ such that $N \leq x < N+1$.

$Proof$ (Existence only). Since $x := \lim_{n\rightarrow \infty}a_n$ and $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence, and for every $a_j$ there exists a integer $N_j$ satisfying $N_j \leq a_j < N_j + 1$. Notice that $|N_j - N_k| \leq |N_j - a_j| + |a_j - a_k| + |a_k - N_k|$, $0 \leq |N_j - a_j| < 1$ for every $j, k \geq 1$, and $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence, we know that for every $\delta > 0$, there exists a integer $M > 0$ such that $|N_j - N_k| < 2 + \delta$ for every $j, k \geq M$. Since $N_j$'s are integers, there are three possible values for $|N_j - N_k|$ : $0, 1, 2$ for every $j, k \geq M$.
It is easy to show that $|N_j - N_k|$ cannot equal to $2$ for every $j, k \geq M$, and the case $|N_j - N_k| = 0$ for every $j, k \geq M$ holds trivial, we only consider the case for which $|N_j - N_k| = 1$ for every $j, k \geq M$. Fix some $k_0 \geq M$ we have $|N_j - N_{k_0}| = 1$ for every $j \geq M$. Rather, we have either $N_j = N_{k_0} + 1$ or $N_j = N_{k_0} - 1$ for every $j \geq M$. Set $N_j = N_{k_0} - 1$ we have $N_{k_0} - 1 \leq a_j < N_{k_0}$ for every $j \geq M$. Let $N := N_{k_0} -1$ we proved $N \leq x < N+1$.          $\Box$
This proof seems not natural enough. It is clear that this proposition is obviously true, but to prove it, we have to deal with many low-level concepts. Is there a neat and clear proof?  Or can we prove that the sequence $\{N_j\}_{j=1}^\infty$ is a Cauchy sequence?
 A: How about:
Lets define sets 
$$L_x=\left \{ x' \in \mathbb{Z} : x' \leq x \right \}$$
$$U_x=\left \{ x' \in \mathbb{Z} : x' > x \right \}$$
As we know, that there exists arbitrary large (and small) integer, we may assume, that for any fixed $x \in \mathbb{R}$, $L_x,U_x \neq \emptyset$
Now let us define $S_x = \sup L_x$ and $I_x =\inf U_x$. As $U_x$ is bounded from below (and symetricaly $L_x$ from above), $S_x= \max L_x \in L_x$ and $I_x = \min U_x \in U_x$. It remains to show, that $S_x+1 = I_x$.
It is quite obvious that $S_x \leq I_x$, so lets assume that $S_x + k = I_x$ and $k>1$, then number $x^*=S_x+(k-1)$ is clearly an integer, and it can be bigger or smaller (or equal) then $x$.


*

*$x^* > x$, then by the definition $x^* \in U_x$, but it is also smaller then $S_x+k=I_x$, while $I_x = \min U_x$, contradiction.

*$x^* \leq x$, thus $x^* \in L_x$, but it is also bigger then $S_x$, while $S_x = \max L_x$, contradiction.


As a result, $S_x + 1 = I_x$ (as definition implies that $S_x \neq I_x$), qued.
Uniquness of the solution is trivial, if there is $N$ and $M$such that $N \leq x < N+1$ and $M \leq x < M+1$ and we assume $N<M$ or $M<N$ then it leads to constradiction with the inequalities provided.
