How can I end my proof about a solution of $x^2-2=0$ in $\mathbb{R}$? In the class we have constructed the real numbers using Dedekind Cuts. Now an exercise ask me to prove that the equation $x^2-2=0$ has a solution in $\mathbb{R}$ using The Least Upper Bound Property. This is my try:
Let $$S=\{ r \in \mathbb{R} \mid r^2<2, r>0 \},$$
so $S \ne \emptyset$, because $1 \in S$, and has an upper bound, for instance 2, then $S$ has least upper bound $s=\sup S$. Now we will prove that $s^2=2$. 
First suppose that $s^2<2$, then $s\in S$ and $s=\max S$. Besides 
$$S=\{r\in \mathbb{R} \mid r^2 \leq s^2, r>0\}=\{r \in \mathbb{R} \mid 0<r\leq s\},$$
then with $$A=\{ r \in \mathbb{R} \mid r\leq 0 \} \cup S,$$ and $$B=\{ r\in \mathbb{R} \mid r\geq s \},$$
we have that $A \cap B = \emptyset$, $\mathbb{R}=A \cup B$ and every element of $A$ is less than every element of $B$. Also $s=\max A$ and $s=\min B$, this is a contradiction (using a a theorem proved in class).
Now suppose that $s^2>2$, and I get stuck here!
Can you check my proof so far? is it good? how can I end it?. Thanks.
 A: As mentioned in comments your reasoning is wrong. Even if you define set $B$ properly so that $A \cap B = \emptyset$ then also I am not sure if we are heading somewhere.
We need to proceed in a different fashion. Suppose $s^{2} < 2$. We also know that $S$ is bounded above by $2$ hence $s \leq 2$. Also it is obvious that $s > 1$ i.e. $s^{2} > 1$. Hence we get $0 < \epsilon = 2 - s^{2} < 1$. Thus we can now choose a real number $t$ such that $s < t < s + \epsilon/5 < s + \epsilon < 3$. Now it is easy to see that $$t^{2} - s^{2} = (t - s)(t + s) < (\epsilon/5)\cdot 5 = \epsilon = 2 - s^{2}$$ and hence $t^{2} < 2$ and so $t \in S$. But $t > s = \sup S$ and we get a contradiction.
If $s^{2} > 2$ so that $s \notin S$ then we can put $ \epsilon = s^{2} - 2$ and obtain (in a similar manner as explained above) a number $t$ such that $t < s$ and $t^{2} > 2$. But then since $s = \sup S$ and $t < s$ there will at least one number say $r \in S$ such that $t < r < s$ and then we have $r^{2} < 2$ because $r \in S$ and $r^{2} > t^{2} > 2$ and hence a contradiction is achieved.
Therefore we must have $s^{2} = 2$.
