Every subgroup of a quotient group is a quotient group itself

You're nearly there. Just use the set-theoretic fact that if $f : S \to T$ is a surjective function, and $B \subseteq T$, then $$ f( f^{-1}(B)) = B. $$

In your case, you obtain that $B = \pi(A) = A/N$.

You were almost done. Since $\pi$ is onto, you have $$A/N=\pi (A) = \pi (\pi^{-1}(B))=B$$