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”!)
Best Answer
Let $X$ be the space of countable ordinals with the order topology (a locally compact Hausdorff space, completely normal, but not metrizable), and let $Y$ be the set of all isolated points of $X$.
Every subset of $Y$ which is closed in $X$ is finite, since every infinite subset of $X$ has a limit point in $X$. (An infinite set of ordinals contains an increasing sequence; the limit of an increasing sequence of countable ordinals is a countable ordinal, i.e., an element of $X$.
$Y$ is not contained in any compact subset of $X$ because no uncountable subset of $X$ is compact or even Lindelöf.
P.S. Here is another example, a first countable, separable, locally compact Hausdorff space $X$ with a dense open subset $Y$ such that: $Y$ is countable and discrete; the only subsets of $Y$ which are closed in $X$ are the finite sets; and $Y$ is not contained in any countably compact subset of $X$.
Let $\mathcal A$ be an infinite maximal almost disjoint family of infinite subsets of $\omega$. $\mathcal A$ must be uncountable, as there is no maximal almost disjoint family of cardinality $\aleph_0$. Let $X$ be the corresponding $\Psi$-space, that is, $X=Y\cup\mathcal A$ where $Y=\omega$, and a set $U\subseteq X$ is open if $A\setminus U$ is finite for each $A\in U\cap\mathcal A$. All the properties claimed above are easily verified; the fact that every infinite subset of $Y=\omega$ has a limit point in $X\setminus Y=\mathcal A$ follows from the maximality of the almost disjoint family $\mathcal A$.
Unlike the previous example, this space $X$ is not normal; if $\mathcal A_0$ is a countably infinite subset of $\mathcal A$, then $\mathcal A_0$ and $\mathcal A\setminus\mathcal A_0$ are disjoint closed sets which can not be separated by open sets.