Consider ω1 equipped with the order topology. Then Borel subsets of ω1 are precisely those which contain a closed and unbounded set or the complement contains such a set. There must be (in ZFC) sets which lack this property, as otherwise ω1 would be measurable. Can one please tell how to “construct” such sets (using some form of choice, of course).
A set S⊆ω1 is stationary if it has non-empty intersection with every cub set, so it suffices to construct two disjoint stationary sets: neither can contain or be disjoint from any cub set. In fact this post from Andres Caicedo’s blog shows how to construct ω1 pairwise disjoint stationary subsets of ω1.
You may also find the two references in Joel David Hamkins’s answer to this MathOverflow question helpful.