$|G:H|=2$, $K \leq G$ with an element not in $H$. Prove $HK=G$ I know it has something to do with normal subgroups and cosets partitioning, but I don't know how to deal with cosets of $HK$.

Since $H$ has index $2$, it is a normal subgroup. This implies that $HK$ is a subgroup. (In general it is a subset). But $H \subsetneq HK \subseteq G$, where the first inclusion is strict because $K$ has an element not in $H$. Comparing indices it follows that $G=HK$.
