= Solution
{c}
If $A$ has a smallest element $a_0$, any open cover has a member containing $a_0$. This member is either the whole line or a ray $(c,\infty)$ with $c<a_0$, so it already contains all of $A$. Thus a single member suffices and $A$ is <compact>.
If $A$ has no smallest element, the family $\{(a,\infty):a\in A\}$ covers $A$: each point has another point of $A$ below it. A finite subfamily has union $(\min_j a_j,\infty)$, which misses its smallest listed endpoint, itself a point of $A$. No finite subcover exists. Therefore \b[a nonempty subset is <compact> exactly when it has a smallest element] in this <topology>.
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