= Solution
\b[No: there are at most three, including zero and the whole group.] A two-generator ordered <Abelian group> is <torsion-free> and has <rank of an abelian group> at most two. A <convex subgroup of an ordered group> is pure: if $ng\in C$ with $n>0$, then $g$ lies between zero and $ng$ or their negatives, so $g\in C$. Therefore a proper inclusion $C<D$ of convex subgroups increases rational rank: an element of $D\setminus C$ cannot have a nonzero integer multiple in $C$. Convex subgroups form a chain, so a rank-two group permits at most two strict inclusions. This is the bound for <convex subgroups of a finite-rank ordered Abelian group>. It is sharp: <lexicographically ordered group> $\mathbb Z^2$ has exactly $0$, $\{0\}\times\mathbb Z$, and $\mathbb Z^2$.
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