 Solution

When $[G:H]=2$, there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$; for $g\in H$, both equal $H$. This is the index-two subgroup is normal argument: $gH=Hg$ for every $g$, which is equivalent to $gH{g}^{-1}=H$. Therefore every subgroup of index two is normal.

In the dihedral group $D_{2n}=⟨r,s⟩$, the rotation cyclic subgroup $⟨r⟩$ has order $n$ and index two, so it is a normal subgroup. It is nontrivial and proper for the usual dihedral convention $n\ge 2$ (polygon symmetries have $n\ge 3$). If the degenerate convention $D_2=C_2$ were allowed, the claimed nontrivial proper subgroup would not exist; the rotation argument uses $n>1$.

For any $k\ge 3$, let $G=S_k$ and take the stabilizer subgroup $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, soConjugation by $(12)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(23)$ belongs to $H$ but does not fix $2$. Thusgives a nonnormal subgroup of every prescribed index $k\ge 3$.