 Solution

There are $2^{{\aleph }_0}$ distinct irrational cuts of the named rational order. Each defines a consistent complete one-type as above, already over the empty additional parameter set, since the rational constants are part of the language. Different cuts cannot be realized by the same element: a named rational between two cuts gives contradictory required inequalities.

An ${\aleph }_0$-saturated model must realize every one of these types, so it has at least $2^{{\aleph }_0}$ Elements. Consequently there is no countable ${\aleph }_0$-saturated model of $T$. This does not contradict existence of a saturated model from the preceding construction, whose size was unrestricted, or existence of the countable prime model, whose requirement is different. No continuum-hypothesis assumption is involved.