real analysis - Prove the existence of the square root of $2$.

I am trying to prove the existence of the square root of $2$. I have some steps with a very vague explanation and I would like to clarify.

The proof: Let

$$S=\{x\in\mathbb R\mid x\geqslant 0 \text{ and } x^2<2\}.$$

I understand the proof of LUB, ∝ and so I am at the step where $\alpha^2=2$.

I know that we are to prove by contradiction so we state let $\alpha^2 <2$ and $\alpha^2 >2$.
Now my instructor wants us to use the Archimedean Axiom $1/n = \varepsilon$.

$(\alpha^2 + 1/n)^2$ then what.....

Answer

Let $E$ be the set $\{y \in R : y\ge 0 \text{ and } y^2 < 2\}$; thus $E$ is the set of all non-negative real numbers whose square is less than $2$. Observe that $E$ has an upper bound of $2$ (because if $y > 2$, then $y^2>4>2$ and hence $y\in E$). Also, $E$ is non-empty (for instance, $1$ is an element of $E$). Thus by the least upper bound property, we have a real number $x:=\sup(E)$ which is the least upper bound of $E$. Then $x$ is greater than or equal to $1$ (since $1\in E$) and less than or equal to $2$ (since $2$ is an upper bound for $E$). So $x$ is positive. Now we show that $x^2=2$.

We argue this by contradiction. We show that both $x^2<2$ and $x^2>2$ lead to contradictions. First suppose that $x^2<2$. Let $0<\epsilon<1$ be a small number; then we have $$(x+\epsilon)^2=x^2+2\epsilon x+\epsilon^2\le x^2+4\epsilon+\epsilon=x^2+5\epsilon$$ since $x\le2$ and $\epsilon^2\le\epsilon$. Since $x^2<2$, we see that we can choose an $0<\epsilon<1$ such that $x^2+5\epsilon<2$, thus $(x+\epsilon)^2<2$. By construction of $E$, this means that $x+\epsilon\in E$; but this contradicts the fact that $x$ is an upper bound of $E$.

Now suppose that $x^2>2$. Let $0<\epsilon<1$ be a small number; then we have $$(x-\epsilon)^2=x^2-2\epsilon x+\epsilon^2\ge x^2-2\epsilon x\ge x^2-4\epsilon$$ since $x\le2$ and $\epsilon^2\ge0$. Since $x^2>2$, we can choose $0<\epsilon<1$ such that $x^2-4\epsilon>2$, and thus $(x-\epsilon)^2>2$. But then this implies that $x-\epsilon\ge y$ for all $y\in E$. (If $x-\epsilon