Compact sets through topologies over a set

Question

I have to show that a set is compact on the Sorgenfrey line. I can prove that it is indeed compact on the usual topology over the real line and my idea is to say that, since the Sorgenfrey topology is finer than the usual topology then it is also a compact of the Sorgenfrey line, but I don't know if this is true and can't prove it.

Answer 1

No, it is not true. Note that $[-1,1]$ is compact in $\mathbb R$ endowed with its usual topology, but it is not compact in $\mathbb R$ endowed with the discrete topology, which is finer than the usual one. If you replace a topology with a finer one, then you get fewer compact sets, not more.

Answer 2

It is shown here that every compact subset of the Sorgenfrey line is at most countable.

It is true that a Sorgenfrey compact subset is Euclidean compact (because a Euclidean open cover is a Sorgenfrey open cover too and has a finite subcover), while going to a finer topology gives us more open covers to find finite subcovers for, so we can lose (as we do here) lots of compact Euclidean compact sets that are no longer compact when seen as subspaces of the Sorgenfrey line.