= Solution
In any <topological space> a closed subset of a <compact> set is <compact>: add its open complement to an open cover and take a finite subcover of the <compact> set. However, <compact> subsets need not be closed in a general space. In the indiscrete two-point space a singleton is <compact> but not closed. Conversely a closed subset need not be <compact>: $\mathbb R$ itself is closed in $\mathbb R$ but the cover $\{(-n,n):n\ge1\}$ has no finite subcover.
In a <Hausdorff space> every <compact subset> $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; <compactness> selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply <compactness>, as the same real-line example shows. It does imply <compactness> if the ambient space is <compact>.
In a <metric space> <compactness> implies <boundedness>: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is <bounded>, but its singleton cover has no finite subcover. It is even closed and <complete>, so closedness plus <boundedness> alone does not suffice in general <metric spaces>.
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