 Solution

Use the constructible universe theorem that $L$ satisfies $ZFC$ and GCH. In particular, internally it has a bijectionThis bijection belongs to $L$ and is also an actual bijection in the ambient universe, because its domain, range and graph are transitive-model sets with the same membership relation.

We must compare the ordinals rather than assume cardinality is absolute. Every ordinal below ${\omega }_1^L$ is countable in $L$, and its witnessing enumeration belongs to the ambient universe too. Thus every such ordinal is countable externally, proving ${\omega }_1^L\le {\omega }_1$. Consequently the ambient cardinality of the constructible reals is at most ${\aleph }_1$.

If every subset of $\omega$ were constructible, the displayed bijection would enumerate the full ambient power set using at most ${\aleph }_1$ points. This contradicts the assumption $2^{{\aleph }_0}={\aleph }_2$. ThereforeThe argument permits ${\omega }_1^L$ to be strictly smaller than the ambient ${\omega }_1$, exactly the cardinal-absoluteness issue highlighted in the source.