Use every non-abelian group of order 6 has a non-normal subgroup of order 2 to classify groups of order 6. 
Prove that every non-abelian group of order $6$ has a non-normal subgroup of order $2$. Use this to classify groups of order $6$.

I proved that every non-abelian group of order 6 has a nonnormal subgroup of order 2, but then how can I use this to classify groups of order 6?

Let $G$ be a non-abelian group of order six.  Then $G$ has exactly $3$ $2$-Sylow subgroups, which map to each other under conjugation.  Denoting these Sylow subgroups $P_1,P_2,P_3$, we obtain a map (the permutation representation of the group action)
$$\varphi: G \to \mathrm{Sym}(P_1,P_2,P_3)\cong S_3,$$
in which $\varphi(g)$ denotes the permutation of $P_1,P_2,P_3$ attained by conjugating each (respectively) by $g$.  This map is injective, hence an isomorphism by orders.  That is, $G \cong S_3$.
For the abelian case, it suffices to prove that such a group $G$ has an element of order $6$.  To do this, it suffices to find elements of orders $2$ and $3$, which exist by a theorem of Cauchy (or later, by the Sylow theorems).
