Are you sure? This proof doesn't use any properties of subgroups or G being Abelian, which is why I'm worried.
I'm not too sure that the implication in
$(e_H, k_2) \in H \times K_2 = H \times K_1$, which implies $k_2 \in K_1$
is true.
The problem: Suppose G is Abelian with two representations as the internal direct product of subgroups: G=HxK1, G=HxK2. Assume K1 is a subset of K2 and show K1=K2.
My attempted solution: I took the element (e_H, k_2), where e_H is the identity element of H and k_2 is an arbitrary element in K2...