Clearly since $|G|<\infty$ and $H\leqslant G$ we have that $|H|<\infty$. Let $h_0\in H$ be fixed and define $\theta_{h_0}:H\to H$ by $h\overset{\theta_{h_0}}{\longmapsto}h_0h$. Clearly this is injective since $\theta_{h_0}(h)=\theta_{h_0}(h')\implies h_0h=h_0h'$ and since $G$ is a group we have cancellation and thus $h=h'$. But, every injection from a finite set to itself is a bijection. Thus, since $e\in H$ we have that $\theta_{h_0}(h)=h_0h=e$ for some $h\in H$. It follows that $h=h_0^{-1}$ and since $h_0$ was arbitrary the conclusion follows.