If $H$ and $G/H$ are compact, then $G$ is compact.

The quotient map $f: G \rightarrow G/H$ is perfect: it is continuous, closed (as $H$ is compact, as you say) and has compact fibres (inverse images of points are compact); the latter is clear as all fibres are homeomorphic to $H$.

And the inverse image of a compact set under a perfect map is compact. And $G$ is the inverse image of $G/H$, which is compact by assumption.

The proof of the latter is not too hard (some details left for you):

Suppose $f: X \rightarrow Y$ is perfect, and let $K \subset Y$ be compact. Let $U_i, i \in I$ be an open cover of $f^{-1}[K]$. For each $k \in K$, we can cover $f^{-1}[\{k\}]$ by finitely many of the $U_i$, call the union of these $U(k)$. Then $O(k) = Y\setminus f[X \setminus U(k)]$ is an open neighbourhood of $k$ in $Y$ (this uses $f$ is a closed map), so finitely many of them cover $K$. The corresponding $U(k)$, and thus the finitely many $U_i$ that compose them, are the required finite subcover of $f^{-1}[K]$.

Another idea is to use that $G/H \times H$ is also compact, and the product (the group product) map is probably (haven't checked the details) a map onto $G$ from this space. This is maybe more appropriate for a course in topological groups...