The property says
If $S$ is nonempty and bounded above, $\sup S$ exists.
Now Rudin is proving
If $S$ is nonempty and bounded below, $\inf S$ exists.
He does this by noticing that the set $L$ of lower bounds of $S$ is nonempty and bounded above, so $\sup L$ exists. Can you see why $\sup L=\inf S$?
Here I prove the claim made in a previous answer, that the set
$$
\mathcal{S} = \left\{x\in\mathbb{Q} : x^2<2 \right\}
$$
does not have a least upper bound in $\mathbb{Q}$.
Since $\mathbb{Q}\subseteq\mathbb{R}$, $\mathcal{S}$ is a subset of $\mathbb{R}$. Furthermore, $\mathcal{S}$ is non-empty (e.g. $1^2 < 2$) and is bounded above by definition, so by the completeness axiom it has a least upper bound in $\mathbb{R}$ - say $\sup\mathcal{S}=s$. We now show that $s = \sqrt{2}$.
There are only 3 possibilities: $s>\sqrt{2}$, $s<\sqrt{2}$, or $s=\sqrt{2}$.
The first possibility can be eliminated from the definition of $\mathcal{S}$. Clearly $\sqrt{2}$ is an upper bound of $\mathcal{S}$, since $x^2 < 2$ implies $x < \sqrt{2}$. Therefore any number larger than $\sqrt{2}$ cannot be the least upper bound.
To eliminate the second possibility, assume for contradiction that $s<\sqrt{2}$. Since the rationals are dense in $\mathbb{R}$, there is a rational $q$ such that $s<q<\sqrt{2}$. But this implies $q^2 < 2$ and so $q\in\mathcal{S}$, which means that $s<q$ cannot be an upper bound of $\mathcal{S}$.
We are left with $s=\sqrt{2}$. Since $\sqrt{2}\notin\mathbb{Q}$, $\mathcal{S}$ does not have a least upper bound in $\mathbb{Q}$ and so we have found a counterexample which shows that $\mathbb{Q}$ is not complete.
Best Answer
They are all wrong.
(A) $\{x\in \mathbb Q\mid x^2<2\}$ is bounded from above (e.g. $42$ is an upper bound), but has no rational number as supremum
(B) $\emptyset$ is bounded from above (e.g. $42$ is an upper bound), but has no real number as supremum (of course this example also works for (A))
(C) Least upper bound property means that every nonempty set that is bounded from above has a least upper bound (in the set). As the exmaple to (A) shows, $\mathbb Q$ does not have this property
(D) $\mathbb R$ is specifically constructed to have the least upper bound property!