 Solution

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. A left coset is $gH=\{gh:h\in H\}$ and a right coset is $Hg=\{hg:h\in H\}$.

If $|G|=2|H|$, then for $g\in H$ both cosets equal $H$. For $g\notin H$, neither coset intersects $H$: an intersection would imply $g\in H$. Both have $|H|$ elements, because multiplication by $g$ is a bijection. They therefore both equal $G\setminus H$, proving $gH=Hg$ in this case too. This establishes that an index-two subgroup is normal.

For Lagrange's theorem, left cosets partition $G$. Indeed if $g_1{h}_1=g_2{h}_2$, then $g_{1}H=g_{2}H$ by multiplying by elements of $H$; otherwise the cosets are disjoint. Every coset has $|H|$ elements, so if there are $[G:H]$ cosets,The group order of $H$ thus divides the group order of $G$. Normality is not needed for this divisibility statement.