When does every compact $A$ in a topology satisfy $A⊆B⊆C$ for some open $B$ and compact $C$? Let $X$ be a topological space such that for any compact subset $A$ of $X$, there exists open set $B$ and compact set $C$ such that $A\subseteq B\subseteq C$. Does this property have a name? If so, what is it? Does this property hold for all topologies? If so I would like a proof and if not a counter example. Thanks.
 A: I don't know if the property has a name, but it doesn't have to hold even for metric spaces. Consider $X=\mathbb{Q}$ with the Euclidean topology and let $A=\{0\}$. Clearly no open subset of $\mathbb{Q}$ is relatively compact (which is the same claim for metric, or even Hausdorff, spaces).

Lemma. If $X$ is Hausdorff then the property is equivalent to local compactness.

Proof. Note that for Hausdorff spaces the property "$V$ is contained in a compact set" is equivalent to "$\overline{V}$ is compact" which is also know as "$V$ is relatively compact".
"$\Rightarrow$" Since $\{x\}$ is compact then by our property it has open neighbourhood $U$ such that $\overline{U}$ is compact. Hence local compactness.
"$\Leftarrow$" Let $A\subseteq X$ be compact. Then for any $x\in A$ there is an open neighbourhood $U_x\subseteq X$ of $x$ that is relatively compact. Since $\{U_x\}_{x\in A}$ cover $A$ then by compactness $A$ is covered by $U_{x_1},\ldots,U_{x_n}$. Clearly $U_{x_1}\cup\cdots\cup U_{x_n}$ is the neighbourhood we are looking for. $\Box$
For non-Hausdorff spaces I suppose we can treat the property as one of the many non-equivalent definitions of local compactness. The name looks appropriate.
A: This condition is a stronger version of one form of local compactness (which considers the case that $A$ is a point); maybe it's equivalent to local compactness under some mild hypotheses, I don't know. In any case it implies local compactness (at least if we also assume that the space is Hausdorff so that all the usual definitions are equivalent), so any non-locally compact space is a counterexample and these are plentiful.
For an explicit counterexample consider any infinite-dimensional normed vector space in the norm topology. By Riesz's lemma we know that the closed unit ball is not compact; this implies that no open subset is contained in a compact subset.
A: The rationals with their usual topology are a counterexample. Let $K$ be a compact subset of $\Bbb Q$. If $U$ is an open nbhd of $K$ in $\Bbb Q$, $\operatorname{cl}U$ is not compact, so $U$ is not contained in any closed subset of $\Bbb Q$. (To see that $\operatorname{cl}U$ is not compact, just note that it contains a non-degenerate interval in $\Bbb Q$.)
A: As Qiaochu Yuan indicated, the property you described requires in particular that each point $x\in X$ have a compact nbhood; that is, the space is weakly locally compact.  It's the weakest form of local compactness, condition (1) in the wikipedia article.
Weak local compactness is also sufficient to satisfy your condition.  To see this, adapt the argument from freakish.  Suppose $A$ is a compact set in $X$.  For each $a\in A$ choose a compact nbhood of it.  The interiors of these compact nbhoods from an open cover of $A$.  So $A$ can be covered by a finite number of these interiors.  And the finite union of the corresponding nbhoods is compact.  That gives you $A\subseteq B\subseteq C$ with $B$ open and $C$ compact.
You can find many examples of spaces that are or are not weakly locally compact in pi-base.  See here in particular.
