Consider $\omega_1$ equipped with the order topology. Then Borel subsets of $\omega_1$ 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 $\omega_1$ would be measurable. Can one please tell how to "construct" such sets (using some form of choice, of course).
A set $S\subseteq\omega_1$ 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 $\omega_1$ pairwise disjoint stationary subsets of $\omega_1$.
You may also find the two references in Joel David Hamkins’s answer to this MathOverflow question helpful.
It requires the axiom of choice, but there are sets which are stationary and co-stationary. Namely they do not contain a club not they a disjoint from one.
For example using Solovay's theorem we can partition $\omega_1$ into $\omega_1$ disjoint stationary sets.
In some models where the axiom if choice fails every subset of $\omega_1$ contains a club or is disjoint from one.