= Solution
Let $F$ be a <closed subset> of the <compact space> $X$. Express an arbitrary <open cover> of $F$ in its <subspace topology> as $\{F\cap U_i:i\in I\}$, where the $U_i$ are <open sets> of $X$. Since $X\setminus F$ is <open>, the family
$$
\{U_i:i\in I\}\cup\{X\setminus F\}
$$
is an <open cover> of $X$. Take a <finite subcover> and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded <set> contains no point of $F$. Their traces on $F$ give the required <finite subcover>. Hence \b[every <closed subset of a compact space> is <compact>], with no <Hausdorff> assumption.
Back to article page
