This question is similar to this previous question.
Let $C\subseteq \mathbb R$ denote the Cantor set and let $A\subseteq C$ be dense with $|A| = |\mathbb R|$, and assume $A$ doesn't contain its infimum. Is there a locally compact, second countable, Hausdorff topology on $A$ that is finer than the right order topology (the topology with base $\{(a,\infty)\cap A: a\in A\}$)? If it helps, the continuum hypothesis can be assumed.
An equivalent formulation of the problem is to find a compact metric on $A\cup \{0\}$ refining the right order topology by identifying the point 0 with the point added in the 1-point compactification.
If $A=C\setminus\{0\}$, the result is trivial as $C$ is compact in the Euclidean topology. If $C\setminus A$ is countable, then the same methods as this answer should hold. The difficulty is if $|C\setminus A| = |\mathbb R|$.
In the Euclidean topology, $A$ is perfect, metrizable, second-countable, nowhere-dense, and hence totally disconnected. We would be done if $A$ were locally compact, so we can assume otherwise. In the right-order topology, $A$ is definitely not Hausdorff (but it is $T_0$), and it is locally compact (neighbourhood base of compact sets), hyperconnected, ultra-connected, and second countable.
There cannot be a compact topology on $A$ that refines the right-order topology, as a set is compact in the right-order topology iff it has the smallest element, and $A$ does not have the smallest element.
The point 0 then feels special. By the same argument, any compact set in the refined topology must have a smallest element (and not necessarily a largest element like it does in the Euclidean topology).
Cross-posted to Math overflow. MO SOLVED the problem
