Let $A$ be the set of all positive rational numbers, and let $B$ be the subset of $A$ defined by $B = \{x \in A \ : \ x^2<2\}$. Does $B$ contain a greatest element?
From my understanding $B$ does contain a supremum, ie a least upper bound. This is because there is a number $x$ such that $x$ is an upper bound of $B$, and any number smaller than $x$ fails to be an upper bound, here $x$ would be $\sqrt2 = 1.41\!\dots$.
What I can't seem to work out is if $B$ has a greatest element?
On
It's well known that $\sqrt{2}$ is irrational, so your $x$ not in $A$. (A better way to say that without referring to the square root of $2$ is to say that there is no rational number whose square is $2$.)
$B$ is bounded above: everything in it is less than $100$, even less than $1.415$. But $B$ has no greatest element in $A$. It does have a supremum $\sqrt{2}$ in the full set of real numbers. That supremum is not a member of $A$, so certainly not a member of $B$.
Hint: If $B$ had a greatest element $y$, then $y$ would be the supremum of $B$ (why?). So if you can prove your supremum $x=\sqrt{2}$ is not itself an element of $B$, that shows that $B$ has no greatest element.
(Note also that it's not quite correct to say that $B$ "contains" a supremum, since that would mean the supremum is an element of $B$. Instead the usual phrasing would be that $B$ "has" a supremum, which does not say anything about whether the supremum is an element of $B$.)