= Solution
If $m=\min A$, any <open cover> contains a member containing $m$. Such a member is either $\mathbb R$ or a ray $(a,\infty)$ with $a<m$. It therefore contains every element of $A$, so that single member is already a <subcover>.
Conversely suppose $A$ is nonempty and has no smallest element. Then
$$
\{(a,\infty):a\in A\}
$$
is an <open cover> of $A$: for any $x\in A$, the lack of a minimum gives an $a\in A$ below $x$. A nonempty finite subcollection has <union> $(a_0,\infty)$, where $a_0$ is the smallest of its thresholds. It misses $a_0\in A$. The empty subcollection cannot cover nonempty $A$ either. Hence there is no finite <subcover>, and $A$ is not <compact>. Thus \b[<compactness> is equivalent to having a smallest element].
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