 Solution

The continuum hypothesis holds. Let $A$ be the assumed Lusin set of cardinality $\mathfrak{c}=2^{{\aleph }_0}$. If $\mathfrak{c}>{\aleph }_1$, choose $B\subseteq A$ with $|B|={\aleph }_1$. The size assumption makes $B$ a meagre set, but then $A\cap B=B$ is uncountable, contradicting the Lusin set property. ThusOne can also use (d) directly: well-order $A$ in type $\mathfrak{c}$. Each proper initial segment has size below $\mathfrak{c}$, so is meagre and consequently countable, since it is a subset of the Lusin set. These initial segments form a linear increasing union of countable sets equal to $A$. The bound in (d) gives $\mathfrak{c}\le {\aleph }_1$.