 Solution

The equivalence classes of an equivalence relation are nonempty, pairwise disjoint, and partition $\mathbb{N}$. To construct a transversal of a set family for them, choose the least member of each class. Its set isEvery class has a least element by the well-ordering principle for the natural numbers, and that element satisfies the displayed condition. A nonleast element fails it because a smaller equivalent element exists. Therefore $S$ meets every class in exactly one point, not merely infinitely many classes.

The complement of $\sim$ is semidecidable, so for a given $n$ run the finitely many semidecision procedures for $m\nsim n$, $0\le m<n$, in parallel. Accept $n$ once all have accepted. If $n$ is a least representative, all finitely many computations halt; otherwise at least one never accepts. For $n=0$, the empty list of tests succeeds immediately. This proves that $S$ is semidecidable. Equivalently, enumerate inequivalent pairs and output $n$ once all pairs $(m,n)$ with $m<n$ have appeared, dovetailing the requirements over all $n$. This is the semidecidable least-representative transversal of a co-computably enumerable equivalence relation. Infinitely many classes make $S$ infinite, but that hypothesis is unnecessary for the existence and semidecidability of a complete transversal.