 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. Thenis 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 compactness is equivalent to having a smallest element.