Every compact subset of a Bernstein set is countable

Suppose, for contradiction, that $K$ is uncountable. Since $K$ is compact, [L1] makes it closed. Let $C$ be the set of condensation points of $K$, that is, the points $x\in K$ for which every neighbourhood of $x$ meets $K$ in uncountably many points. For each $x\in K\setminus C$, take the first rational open interval in a fixed enumeration that contains $x$ and meets $K$ countably. Thus $K\setminus C$ is covered by a countable family of countable intersections with $K$, and [L5] makes it countable. Hence $C$ is uncountable and nonempty.

The set $C$ is perfect. It is closed, because if $x\mathrm{\notin }C$ then some neighbourhood of $x$ meets $K$ only countably, and that same neighbourhood avoids $C$; and it has no isolated points, because every neighbourhood of a point of $C$ meets $K$ uncountably, while $K\setminus C$ is countable by step 1.1, so the same neighbourhood meets $C$ in a point different from the centre. Thus [L4] applies to $C$.

Now $C\subseteq K\subseteq B$, so the nonempty perfect set $C$ misses the complement of $B$, contradicting [F1]. Therefore the assumption in step 1.1 was false, and $K$ is countable.