my guess would be that G is cyclic. let's see if we can prove this.
suppose H is such a group. since H is proper, there must be some x in G, with x not in H.
now <x> is a subgroup of G. if it were proper, then by our assumption on H, <x> would be contained in H (since H contains every proper subgroup).
in particular, x would have to be in H, since x is in <x>. but this contradicts our choice of x.
hence <x> must not be proper, in other words, <x> = G, so G is cyclic.