if $H \leq G$ has index 2, then $a^2\in H$ for every $a\in G$ if $H \leq G$ has index 2, then $a^2\in H$ for every $a\in G$
I am not sure that whether the way that i prove this statement is correct.
Since $[G:H]=2, \forall a\in G,G/H=\{H,Ha\}$
Hence $Ha^2=H \implies a^2\in H$

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 $a^2\in H$.
