= Solution
{c}
In the <subspace topology> on $S$, a subset $A$ is <open> in $S$ precisely when $A=S\cap O$ for some <open set> $O\subseteq\mathbb C$. A subset $B$ is <closed> in $S$ precisely when $S\setminus B$ is open in $S$, equivalently $B=S\cap F$ for some <closed set> $F\subseteq\mathbb C$.
A <compact space> is one for which every <open cover> has a finite <subcover>. Here the cover may use relatively open subsets of $S$, or equivalently ambient open subsets covering it. The <Heine-Borel theorem> states that
$$
\boxed{S\subseteq\mathbb C\text{ is compact}\iff S\text{ is closed and bounded}.}
$$
Closedness and boundedness here refer to the ambient <complex plane>, not merely closedness in $S$ itself.
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