= Solution
In an arbitrary <topological space>, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb R$ is closed in itself but the cover $\{(-n,n):n\geq1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a <compact space> $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.
In a <Hausdorff space>, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint <open set>[open sets] $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.
In every <metric space>, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the <triangle inequality> bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus \b[compact implies closed and bounded in a <metric space>, while closed and bounded need not imply compact].
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