Here is a question on the book "Problem -Solving Strategies" by Arthur Engel, with a provided solution:

In the solution, they partitioned the abelian group into 3 sets such that none of them is sum-free. However, in $A_3$, $(a+b+d) +(a+c+d)=b+c \in A_3$.

You are right. To correct the proof, move $a+c+d$ to $A_2$ and $b+c+d$ to $A_3$. Is everything all right now?