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.
Suppose that there's at least two integers such that the property holds, and suppose that $M<N$.
Now, $M<N \le x <M+1<N+1$ according to the property, but we have that $M<N<M+1$ , but clearly because they're integers (is there an integer $N$ such that there is integers $M$ and $M+1$ and $M<N<M+1$ holds) , $M=N$ should also hold and we have a contradiction.
Best Answer
What you are asking for is the existence of $n$-th root. This is true in $\mathbb{R}$ but not in $\mathbb{Q}$ (for example $\not \exists q\in \mathbb{Q}:q^2=2$). Any proof of this will require the completeness of the real line (Least Upper Bound Property).
LUB property: Every subset of $\mathbb{R}$ bounded above must have a supremum.
Here is a typical proof of this theorem only using the LUB property and the Binomial Theorem.
Let $x>1$ and $S=\left\{ a>0: a^n<x \right\}$. Obviously $1\in S$ and so $S\neq \emptyset $.
Since $x>1\Rightarrow x^n>x$ we have that $S$ is bounded above by $x$. Therefore, by the Least Upper Bound Property, $\exists \sup S=r\in \mathbb{R}$.
We shall prove that $r^n=x$.
Suppose that $r^n>x$ and let \begin{equation}\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{r^n-x}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}\end{equation} Then $0<\epsilon <1$ and so $\epsilon ^n<1$.
Therefore, by the Binomial Theorem, \begin{gather}(r-\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}(-\epsilon )^k}=r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}(-1)^{k-1}\ge \\ r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}> r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r-\epsilon)^n> r^n-\dfrac{r^n-x}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}=r^n-r^n+x=x\Rightarrow \left( r-\epsilon \right)^n>x\end{gather} which is a contradiction since $r=\sup S$ and $r-\epsilon <r$
Suppose that $r^n<x$ and let \begin{equation}\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{x-r^n}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}\end{equation} Then, $0<\epsilon <1$ and so $\epsilon ^n<1$. Therefore, by the Binomial Theorem, \begin{gather}(r+\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}\epsilon^k}=r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}<\\ r^n+\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\ (r+\epsilon)^n< r^n+\dfrac{x-r^n}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}= r^n+x-r^n=x\Rightarrow (r+\epsilon)^n<x\end{gather} and so $\sup S=r<r+\epsilon\in S$ which is a contradiction.
Therefore, $r^n=x$ if $x>1$
Let $0<x<1$. Then $\frac{1}{x}>1\Rightarrow \exists r>0:r^n=\frac{1}{x}\Rightarrow \exists r'=\frac{1}{r}>0:r'^n=\frac{1}{r}^n=\frac{1}{r^n}=x$.
If $x=1$ then $r=1$.
EDIT: Motivation as per request: As I said this is statement is not true if we replace $\mathbb{R}$ with $\mathbb{Q}$. What sets $\mathbb{R}$ and $\mathbb{Q}$ apart is the completeness of $\mathbb{R}$. The LUB property therefore must in some way be used. This is why we define $S$. Because the LUB is an existensial theorem, it shows the existence of a supremum but not its value, it is often used in proofs by contradiction.
So assuming $r^n>x$ we need to arrive to a contradiction. Remember $r$ is a very special number, the supremum of $S$. If we could show that $\exists m\in \mathbb{R}$ so that $m<r$ and is an upper bound of $S$ or $m>r$ and $m\in S$ then we are done. This is what we do with $m=r-\epsilon$ in the first case and with $m=r+\epsilon$ in the second.
It all boils down to finding an $\epsilon>0$ so that $r-\epsilon$ is an upper bound of $S$, that is $(r-\epsilon)^n>x$. We can make this $\epsilon$ as small as we want. I shall choose an $\epsilon$ for $n=2$. \begin{equation}(r-\epsilon )^2=r^2-2r\epsilon+\epsilon^2>r^2-2r\epsilon-\epsilon\end{equation} Remember we want $(r-\epsilon )^2>x$ and so it suffices \begin{equation}r^2-2r\epsilon-\epsilon>x\Leftrightarrow \epsilon<\frac{r^2-x}{1+2r}\end{equation}