 Solution

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 a nonempty subset is compact exactly when it has a smallest element in this topology.