This proves that HB is equivalent to the fact that every bounded subset is a subset of a compact set. Clearly, I am doing it on a metric space, only then this makes sense. As pointed out OP seems to understand this equivalence and hence is irrelevant.
Heine-Borel Property:
A metric space $(X, d)$ is said to be Heine-Borel if any closed and bounded subset of it is compact.
You'd like to know the name of those spaces in which a bounded set is contained in a compact set (BIC).
We claim that this property is equivalent to the Heine-Borel (HB) Property.
- $(BIC\implies HB)$ Suppose a space $X$ is $BIC$. Let $B$ bounded subset of $X$. Because $X$ is $BIC$, $B \subseteq K$ where $K$ is compact. If $B$ is also closed, you have closed subset of a compact set which is compact. Hence the space is $HB$.
- $(HB \implies BIC)$ Let $B$ bounded subset of $X$ that has $HB$. We need to find a compact subset $K$ such that $B \subseteq K$. Then the closure, $\bar B$ is closed and bounded, hence compact, because of $HB$ property and note that $B \subseteq \bar B$. Hence, $X$ is $BIC$.
So, we can as well call these Heine-Borel spaces.
In fact, another characterisation of Heine Borel spaces is that, bounded sets are also totally bounded.