Since the construction of a partition of two Bernstein sets is almost identical to that of a partition of three in an uncountable Polish space. It’s possible that the union of a Bernstein set and a singleton is a Bernstein set. But is it possible otherwise, that is, the union of a Bernstein set and a singleton contains an uncountable closed set ?
To see this, notice that if you remove a point from a perfect set, what remains will still contain a perfect set, so the intersection of a Bernstein set with a perfect set is not just nonempty, but also infinite. In fact, it’s dense (in the perfect set) by a similar argument.
Therefore, removing (or adding) a single point does not turn a Bernstein set to a non-Bernstein set.
Using measure arguments you can show that the intersection is uncountable: the intersection of a Bernstein set with any perfect set has full outer measure with respect to any continuous probability measure on the perfect set, so it can’t be countable.
(I vaguely recall that any Bernstein set is a union of $\mathfrak c$-many Bernstein sets, so that the intersection is actually of cardinality of the continuum, but I don’t remember the proof, exactly, and might be mixing something up.)