If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies $x$ $\le$ $n$ $<$ $x$ $+$ $1$. 
If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies $x \le n < x + 1$.

I've come up with a proof myself, but I'd like you guys to critique it, because I'm not totally convinced that I'm being as rigorous as I could be for an analysis study context.
Let $A = \{ n \in \Bbb Z : n\ge x\}$.
Let $n$ be the smallest integer greater than $x$ [Here's where I most doubt of the rigorousness of my proof. Am I allowed to summon this $n$ out of nowhere?]. Since $n\in A$, $n \ge\inf A$. But, if $n >\inf A$ there is a non-integer real number $y$ such that $\inf A < y < n$, which is absurd, because if it was so $y$ would be a lower bound for $A$ greater than the greatest lower bound. Hence $n =\inf A$.
From the definition of infimum, we have $x \le \inf A$, since any element of $A$ is equal to or greater than $x$. There are, therefore, two possible cases:
i) If $x =\inf A \Rightarrow \inf A < x + 1 \Rightarrow n < x + 1$. Therefore $n$ is an integer which satisfies $x \le n < x + 1$.
ii) If $x < \inf A \Rightarrow x < n$. In this case, $0 < n - x < 1$, because $n$ is the smallest integer greater than $x$, which implies that $n < x + 1$. Therefore $n$ is an integer which satisfies $x \le n < x + 1$. 
About the uniqueness of $n$:
If $m \in A \Rightarrow m \ge n$, because $n = \inf A$. But if $m > n \Rightarrow m \ge n + 1 \ge x + 1$, hence $m$ doesn't satisfy $x \le m < x + 1$. Therefore we conclude that $n$ is unique.
 A: If there were none, there would two consecutive integers (the smallest before $x$, the largest at least $x+1$) which differ by more than $1$.
If there were more than $1$, there would be two integers which differ by less than $1$.
A: I'll use quite the same logic, but will try to be rigorous.
Let's first choose large enough $k \in \mathbb{N}$ such that $x'=x+k>0$ (this exists as $\mathbb{N}$ is unbounded.
Let, $$S = \{ m \in \mathbb{N} \mid m \geq x'\}$$
Now, as $S$ is a set of natural numbers, by well ordering property, it has a least number, say $n'$.
Now, $x'+1 \leq n' \implies x' \leq n'-1$. But, that's not possible, since $n'$ is the smallest element of $S$. Hence, $x'+1 > n'$.
Put $n=n'-k$. Clearly, this is the desired $n$.
The uniqueness part is obvious as $x \leq n,n' <x+1 \implies |n-n'|<1 \implies n=n'$ as both of them are integers.
EDIT : Some really dumb errors corrected after 3 long years thanks to @user0.
A: let $x = [x]+\{x\}$, here $[x]$ is the maximum integer which less than $x$, so $0 \le \{x\} < 1$.
so we have $[x]+\{x\} \le n \le [x]+1+\{x\}$, let $y = n-[x]$, it is $\{x\} \le y \le 1+\{x\}$, just need to proof there is a integer $y$. but clearly, $y=1$ is the solution.
A: First suppose there are more than one integers (namely m > 1 integers) such that 

1) $x \le n \le n+1 \le ... \le n+m-1 < x+1$. 

Subtract x through the inequality 1): 

2) $0 \le n-x \le n+1-x \le ... \le n+m-1-x < 1$. 

This means $n+1-x < 1$. Subtract one: $n-x < 0$. This contradicts the first inequality in 2), hence there aren't more than one integers.
Second suppose there are zero integers such that:

3) $k < x < x+1 \le k+1$

where k is an integer. Now subtract k from 3):

4) $0 < x-k < x+1-k \le 1$

This means that $x+1-k \le 1$. Subtract one: $x-k \le 0$. However this contradicts 4), so there must be one integer.
