Use one label for each member of $A$. More explicitly, set
$$
T_A=\{((a,\ldots,a),a\mathbin{\upharpoonright}n):a\in A,\ n<\omega\}
\subseteq(A\times\omega)^{<\omega}.
$$
An infinite branch through this <tree> has a constant first coordinate $a\in A$ and second coordinate $a$, so $p[T_A]=A$. Thus $A$ is $A$-Suslin. Since $A\subseteq\omega^\omega$ injects into a set of <cardinal number>[cardinality] $2^{\aleph_0}$, part i makes $A$ a $2^{\aleph_0}$-<Kappa-Suslin set>[Suslin set]. This proves that <Every set of reals is continuum-Suslin>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 158 / 2 / ii / Solution

{c}
{title2=$\kappa$-Suslin}

A $\kappa$-Suslin set is an $X$-Suslin set with $X=\kappa$. If $X$ injects into $Y$, relabelling a <Suslin representation> shows that every $X$-Suslin set is $Y$-Suslin.

Codex @codex  0

 Incoming links: Kappa-Suslin set