If $H$ and $G/H$ are compact, then $G$ is compact. Suppose that $G$ is a topological group and that $H$ is  a subgroup of $G$ so that $H$ and $G/H$ are compact. I am trying to show that $G$ must be compact. 
The first idea is to use the natural map $f: G \rightarrow G/H$, this is a closed map since $H$ is compact. Thus if $\mathscr{C}$ is a collection of closed subsets of $G$ with finite intersection property (FIP), then $\{f(C): C \in \mathscr{C}\}$ a collection of closed subsets of $G/H$. This collection also has FIP because 
$$f(C_1 \cap \ldots \cap C_n) \subseteq f(C_1) \cap \ldots \cap f(C_n)$$
so $\{f(C): C \in \mathscr{C}\}$ has nonempty intersection because $G/H$ is compact. So there exists some coset $gH$ that is contained in every $f(C)$, that is $C \cap gH$ is nonempty. My next idea would be to show that $\{C \cap gH: C \in \mathscr{C}\}$ is a collection of closed sets in $gH$ with FIP. Then this collection, and thus $\mathscr{C}$ would have nonempty intersection. 
We know that $C \cap gH$ is closed in $gH$, but I don't know how to show that it has FIP. Right now I think that might not even be true in general, since our choice of $gH$ was arbitrary. How can I choose $gH$ in the intersection $\cap_{C \in \mathscr{C}} f(C)$ so that $\cap_{C \in \mathscr{C}} C \cap gH$ is nonempty? Does this proof idea work?
 A: 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...
A: WLOG we can assume $\mathscr{C}$ to be closed under finite intersection (if not, then just consider  $\overline{\mathscr{C}}$ the extension of $\mathscr{C}$ by adding all finite intersections of $\mathscr{C}$). So, in that case if $\mathscr{F}$ is a finite collection of sets in $\mathscr{C}$, we have that $C'=\bigcap_{C\in \mathscr{F} } C$ is in $\mathscr{C}$ and therefore 
$\bigcap_{C\in \mathscr{F}} (C\cap gH)=gH\cap C'\neq\emptyset$. 
Proving that $\{C\cap gH: C\in\mathscr{C}\}$ has the finite intersection property.
