Notice that the usual construction of the Bernstein set can also yield $\mathfrak c$ disjoint Bernstein sets in an arbitrary uncountable Polish space by working diagonally, similarly to how you construct a bijection between $\mathbf N$ and $\mathbf N^2$: you take one point from the first closed set into the first Bernstein set, two points from what remains of the second closed set to first two Bernstein sets etc, resulting in a sequence of distjoint sets, each of which intersects all but possibly some $\kappa<\mathfrak c$ uncountable closed sets. Then you notice that every (uncountable) closed set contains $\mathfrak c$ other (uncountable) closed subsets, so each of those actually intersects ALL uncountable closed sets.
In many ways, Bernstein sets are "as pathological as possible" for a subset of a Polish space, and as such are often good source of examples of pathological behavior.
Bernstein sets are not only non-measurable (and extremely so: the outer measure of a Bernstein set with respect to any continuous Borel measure is full, while the inner is zero!), but also do not have Baire property.
They are also quite interesting as topological measure spaces. Every compact subset of the Bernstein set is countable, so it's impossible to define a (nontrivial) continuous Radon measure on Bernstein set, which makes them sort of the opposite of Polish spaces -- in a Polish space, any finite Borel measure is Radon.