Proof.
First we discuss the case where G is commutative
or H is normal in G.
Then G/H is a locally compact topological group.
It is well-known that the Haar measure on a locally
compact topological group is finite if and only if this group
is compact.
(This is not only well-known, but also easy to prove: Let A be a non-compact locally compact topological group. Let K be a compact subset of positive Haar measure in A. Then the set of all x in A for which xK and K have non-empty intersection is a compact set and therefore does not equal the whole of A. Hence there is an element x in A such that K and xK are disjoint. The union of K and xK is a compact subset whose Haar measure is the twofold of the measure of K. By iteration it follows that A contains compact subsets of arbitrarily large measure. In particular, the Haar measure of A is not finite.)
Now let us discuss the general case.
Let Z denote the center of G. Then ZH is a subgroup of G. Let I denote the closure of ZH in G. This is again a subgroup of G. We obtain a fibration from G/H to G/I. It follows that both G/I and I/H admit a G- resp. I-invariant probability measure. In order to show that G/H is compact, it suffices to show that both G/I and I/H are compact.
H is a normal subgroup of ZH, because Z is central. Since the normalizer of H is closed, it follows that H is normal in I, i.e. I/H is a topological group. Thus I/H is a locally compact topological group with finite Haar measure and therefore compact.
The base G/I can be realized as the quotient of G/Z by I/Z. Observe that the central series of G/Z is one step smaller than that of G. Furthermore we proved already the statement for commutative G. Thus we may apply induction on the length of the central series of G and thereby complete the proof.