The strategy is to show $gK = Kg$ for all $g \in G$ by considering two cases: $g \in K$ and $g \notin K$. **Step 1: $g \in K$.** If $g \in K$, then $gK = K = Kg$ (since the coset of an element of $K$ is $K$ itself). **Step 2: $g \notin K$.** Since $|G : K| = 2$, there are exactly two left cosets of $K$ in $G$: the subgroup $K$ itself and one other. Since $g \notin K$, the left coset $gK \neq K$, so the partition of $G$ into left cosets is: \begin{align*} G = K \;\dot\cup\; gK. \end{align*} The identical argument for right cosets gives: \begin{align*} G = K \;\dot\cup\; Kg. \end{align*} Since $K$ and $gK$ are disjoint, and $K$ and $Kg$ are disjoint, and both unions equal $G$, it follows that $gK = G \setminus K = Kg$. **Step 3: Conclude.** In both cases $gK = Kg$, so $K \unlhd G$.