= 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 $gHg^{-1}=H$. Therefore \b[every <subgroup> of index two is normal].
In the <dihedral group> $D_{2n}=\langle r,s\rangle$, the rotation <cyclic subgroup> $\langle r\rangle$ has order $n$ and index two, so it is a <normal subgroup>. It is nontrivial and proper for the usual dihedral convention $n\geq2$ (polygon symmetries have $n\geq3$). 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 $k\geq3$, let $G=S_k$ and take the <stabilizer subgroup> $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, so
$$
|G|=k!,\qquad |H|=(k-1)!,\qquad [G:H]=k.
$$
Conjugation by $(1\ 2)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(2\ 3)$ belongs to $H$ but does not fix $2$. Thus
$$
\boxed{G=S_k,\quad H=\{\sigma:\sigma(1)=1\}}
$$
gives a nonnormal <subgroup> of every prescribed index $k\geq3$.
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