Or, in other words, prove that an infinite set $S \subset \mathbb{N}$ that has no infinite decidable subsets also has no infinite enumerable subsets.
One idea I had is to show that the complement of $S$ has a non-empty intersection with any infinite enumerable set (kind of the reverse of how the existence of simple sets is proven), but since $S$ is not something we construct, I'm not sure how to proceed.
Self-answering my second question on this branch of math.
I managed to prove that every enumerable set $S$ has a decidable subset (it's sufficient to show a monotonically increasing function defined via a computable function $f$ whose domain or codomain is $S$). The result immediately follows.