It is well known that closed subsets of compact sets are themselves compact. Now the reverse is not true: A set of which all closed subsets are compact needs not to be compact itself; for example, consider non-closed bounded sets in $\mathbb R^n$.
However those sets are themselves subsets of compact sets (as bounded sets, they are subsets of closed balls, which are compact). And it is obvious that the initially quoted theorem also holds for arbitrary subsets of compact sets, since the subset relation is transitive.
However I wonder: Can there exist a set in some topological space, no matter how weird, such that all closed subsets of that set are compact, but the set itself is not the subset of a compact set?
There was a related question that asked about the case where all proper closed subsets of a topological space are compact, and the conclusion was that the space itself is compact. However if this helps with the subset case, then I don't see how.
Clarification: Since it seems to have caused a lot of confusion in the comments: In the context of my post, “closed” is to be understood in the topology of the full space, not in the subspace topology of the subset (those are very different notions of “closed”!)