= Solution
Choose $x\notin H$, possible because $H$ is proper. For any $h\in H$, the element $xh$ is also outside $H$, since otherwise multiplying by $h^{-1}$ would put $x$ inside $H$. Thus both $x$ and $xh$ belong to $K=\langle G\setminus H\rangle$, and $h=x^{-1}(xh)\in K$. So $K$ contains $H$ as well as its entire complement, proving
$$
\boxed{\langle G\setminus H\rangle=G.}
$$
The <complement of a proper subgroup generates the group> argument does not actually require finiteness.
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