= Solution
The <countable> <omitting types theorem> says that for a consistent theory in a <countable> language and countably many nonprincipal types of finite arities, there is a <countable> model omitting all those types. In particular a nonisolated <complete type> of $T$ can be omitted in a <countable> model.
Interpret “definable” here as \b[without parameters]. If parameters were allowed freely, every infinite structure would have infinitely many definable singleton sets, and the claimed implication would be false even for a <dense linear order without endpoints>.
Infinitely many distinct parameter-free definable subsets of $A^n$ imply infinitely many complete $n$-types: if there were only $r$ such types, each formula would select a subset of those $r$ types, giving at most $2^r$ distinct definable subsets. There must also be a nonisolated <complete type>. To see this, suppose each <complete type> $p$ had an isolator $\theta_p$. The family of their basic open sets covers the <type space>. Its <compactness theorem> gives a finite subcover, which would account for only finitely many <complete types>, a contradiction. Equivalently apply logical <compactness theorem> to $T\cup\{\neg\theta_p(c):p\text{ complete}\}$.
Choose a nonisolated type $p$. The <omitting types theorem> gives a <countable> model omitting it, while part (d) gives a <countable> model realizing it. An <isomorphism of structures> preserves realization of parameter-free types, so these models are not isomorphic. Thus $\boxed{T\text{ is not }\aleph_0\text{-categorical}}$.
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