The set $$S=\{s\in \mathbb{Q} |1\leq s^2\leq 2\}$$ has no maximal element with respect to the order in $\mathbb{R}$
Suppose it has a maximal element say $m$ then for every chain in S, if $s\in S $ satisfies $m\leq s$ then we must have $s\leq m$
That is if $1\leq s^2 \leq 2$ and $m\leq s$ then $s\leq m$, but no such $m$ exists as then S must have a Supremum which it doesn't have.
Hence, S cannot have a maximal element. I can prove it this way but, Wikipedia has another proof which I'm unable to understand
If $m$ is a maximal element and $s \in S$; then it remains possible that neither $s\leq m$ nor $m\leq s$.This leaves open the possibility that there exists more than one maximal elements.
I didn't get the bold portion. How does it come ?
The phrase you quote from the article in Wikipeda does not refer to Example 2, but to Example 3 which is a non-linear order.
Your argument looks incomplete, because you don't show where you would use that the bound is $2$; if you replace $2$ with $4$, say, the maximum does exist.
Here is a way to make the argument work. Suppose that $m=\max S$. Here we are working in the context of $\mathbb Q$, we are not even mentioning the reals, so we are assuming that $m\in\mathbb Q$. So $m\in S$ and hence $m^2\leq 2$. The point now is to show that $m^2=2$, which will give the desired contradiction. Suppose that $m^2<2$. Note that $m>1$, so $m<m^2$. We have $$ (m+\delta)^2=m^2+2m\delta+\delta^2\leq m^2+2m^2\delta+\delta^2=m^2(1+2\delta)+\delta^2. $$ Choosing $\delta$ small enough, we get $s+\delta\in S$ and $s+\delta>m$, a contradiction. So $m^2\geq 2$. As we already know that $m^2\leq 2$, we get $m^2=2$, the desired contradiction.