Properties of separable metric space

Question

I am reading Appendix M3 in Patrick Billingsley (1999) "Convergence of probability measures" which deals with metric spaces.

The space $S$ is separable if it contains a countable, dense subset.

And then there is the following theorem (p.237)

These three conditions are equivalent:

1. $S$ is separable.
2. $S$ contains a countable base.
3. Each open cover of each subset of $S$ has a countable subcover.

Questions:

1. Am I correct to say that the second condition is only equivalent because $S$ is metric, so second-countable?
2. Why is the third condition stated in terms of subsets of $S$? Is it because, in general, subsets of separable spaces are not separable but it is true when $S$ is metric? I am confused because later on the same page there is another theorem, which starts with

Suppose the subset $M$ of $S$ is separable...

why do we make this assumption (or at least frame it as an assumption) if all subsets of separable metric spaces are separable?

Answer 1

Those three conditions can be named as follows:

1. $S$ is separable,
2. $S$ is second countable,
3. $S$ is hereditarily Lindelöf.

These three notions are equivalent in the class of metric spaces. For example $\Bbb R$ with lower limit topology (Sorgenfrey line) is separable but not second countable.

Yes, subsets of separable spaces don't need to be separable, but subsets of second countable spaces are second countable. Therefore, in the class of metric spaces, subsets of separable spaces are indeed separable.

Regarding the third point, it implies that the space is Lindelöf, which implies that it's second countable (in metric spaces), then it's each subspace is also second countable, so it's Lindelöf.