if $H \leq G$ has index 2, then $a^2\in H$ for every $a\in G$

Your idea is close, but the proof is not entirely correct as stated. Suppose $H$ has index $2$. Let $a\in G$. If $a\in H$, then $a^2\in H$ certainly. If $a\notin H$, then $G/H=\{H,aH\}$. Also, $H$ is normal, so $G/H$ is a group. Now $aH$ has order $2$ in $G/H$, so $$ H=(aH)^2=a^2H, $$ and thus $a^2\in H$.

Note $aH = bH \iff b^{-1}a \in H$. As others have remarked above, if $a \in H$, then certainly $a^2 \in H$, and in this case $a^2H = H$ (since $a^2 = e^{-1}a^2 \in H$).

If $a \not \in H$, we have only two choices for $a^2H$: either $H$, or $aH$. Suppose we somehow had:

$a^2H = aH$.

This implies that $a^{-1}a^2\in H$, that is: $a \in H$, a contradiction.