Prove the existence of the square root of $2$.

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 $x-\epsilon<y$ then $(x-\epsilon)^2<y^2\le2$, a contradiction.) Thus $x-\epsilon$ is an upper bound for $E$, which contradicts the fact that $x$ is the least upper bound of $E$. From these two contradictions we see that $x^2=2$, as desired.

$\sqrt 2$ is the positive root of $x^2-2$.

Let $f(x)=x^2-2$. Then $f$ is continuous and $f(1)f(2)<0$. Now we can deduce that $f(x)$ has a root betwen $1$ and $2$ by IVT.

The statement $x \in S$ is equivalent to $x \geq 0 \text{ and } x^2 < 2$. So in particular $x \in S \Rightarrow x^2 < 2$. The contrapositive of this latter statement is $x^2 \geq 2 \Rightarrow x \not \in S$. Does this answer your question?

You can use $f(x)=x^{2}-2$ then $f(3)=7>0$ and $f(1)=-1<0$, now use Weierstras theorem for continuous functions. for the method that you want, let $\alpha$ be the Lub for then $b=2-\alpha^2>0$ there is $n_0$ such that $b=2-\alpha^2>\dfrac{1}{n_0}$. now it is may choose $n>n_0$ such that $(\alpha+\dfrac{1}{n})^2<2$ contrary that $\alpha$ is Lub

It is no contrapositive statement since it is no statement. A set of numbers is defined with such and such conditions, and it is proved this set is bounded from above by $2$. That's all.