 Countable-family diamond equivalence

Enumerate each countable guessing family and use a bijection $\pi :\lambda \times \omega \to \lambda$ with a club set of prefix-closure points. The $n$th candidate sequence decodes the $n$th component of the $n$th family entry. If all candidates fail, choose counterexample sets $X_n$ and club sets witnessing failure, then code the $X_n$ together using $\pi$. A correct family guess on the common closure club set and all failure club sets decodes to one of the forbidden correct guesses. The singleton-family implication supplies the reverse direction.