I have some doubts in proof.
R is separable, and thus has a countable dense set, namely Q. Let G ⊂ R be any open set.
Then Q∩G is a countable dense set in G by the Archimedean property, and since G is open we can choose an open interval around every rational in G. Then G is the union of that countable collection of intervals. However, we need to find a countable collection of disjoint intervals.
Notice that the union of any intervals which contain the same point is an interval with a lower endpoint equal to the infimum of the lower endpoints of the intervals (possibly −∞) and with an upper endpoint equal to the supremum of the upper endpoints of the intervals (possibly ∞). We create a new countable collection of intervals whose union is G by the following procedure.
Take any point in G ∩Q and take the union of all intervals in G that contain it. Call this interval I1. Now take some point in (G \ I1) ∩Q and take the union of all intervals in G \ I1 that contain it. Repeating this process we get a countable collection of disjoint intervals I1,I2,I3,..., each of which is in G and which together cover G.
Please explain statements in dark / bold letters. Sorry if this topic is repeated but I need to clear this proof