a) Let G:=(A,+) be the additive group structure of A. Fix a∈D. Consider the group endomorphism φ:G→G given by x↦axa. For a unit x, since ax is not a unit, it follows that axax=(ax)2=0 so axa=0. Hence A×⊆kerφ.
Assume FTSOC there exists x∈A∖kerφ. Then x is not a unit. Since a∈kerφ, it follows that a+x∉kerφ so a+x is not a unit. Thus ax+xa=(a+x)2=0 so φ(x)=axa=a(ax+xa)=0, contradicting the choice of x. ◻
b) Let a=a1⋯ak be the longest nonzero product of elements in D. It exists since D is finite, and the ai are distinct by part a). Clearly a∈D since a1a=0. If there exists b∈D such that ab or ba is nonzero, ab or ba would be a longer nonzero product of elements in D, contradicting the maximality of a. Thus ab=ba=0 for all b∈D, so this choice of a works. ◻