How is this different from (b)? If you mean the difference is in $\ge$ vs $>$, then note that your statement of negation obviously implies (b). But suppose (b) holds and you want to prove

\[

\forall C > 0\;\exists n \in \Bbb{N}\; |x_n| > C.\qquad(*)

\]

Fix an arbitrary $C$ and feed $C+1$ to (b) (i.e., replace $C$ in (b) with $C+1$ where $C$ is the value you fixed). It will spit out some $n$ such that $|x_n|\ge C+1$. So $|x_n|>C$ for the same $n$.

It is also trivial that (d) implies (b) and (*). But if (b) holds and you want to prove (d), you fix an arbitrary $C$ and then instantiate the universally quantified $C$ in (b) with $C+1$, $C+2$, $C+3$ and so on. This way you get infinitely many sequence members that exceed $C$.