 Every set of reals is continuum-Suslin

Assume the axiom of choice, and let $A\subseteq {\omega }^{\omega }$. Every such $A$ is $2^{{\aleph }_0}$-Suslin. To see this, use the elements of Baire space of sequences as labels. For each $a\in A$, put the nodeof length $n$ into a tree, including its prefixes. The label sequence repeats the same entire real $a$ at every position. An infinite branch therefore projects to that particular $a$, rather than merely to a limit of elements of $A$. The projection is exactly $A$.

A bijection between the label set ${\omega }^{\omega }$ and a cardinal of size $2^{{\aleph }_0}$ converts this into the usual Kappa-Suslin set representation. The well-ordering of the labels uses choice; without it, the real-labelled tree is still available but an ordinal-sized relabelling needs a separate hypothesis.