Let $T$ be pruned subtree of $\omega^{<\omega}$. For my cases of interest, we may assume that $T$ is infinitely branching at every node, and consists of increasing sequences.

Let $A=\{x\in\omega^{\omega}:\exists y\in[T](x \text{ is a subsequence of } y)\}$, where $[T]$ denotes the (closed) set of infinite branches through $T$. Is $A$ necessarily a Borel set?

Clearly, $A$ is analytic. It might be useful to note the following: Given an infinite increasing $x\in\omega^\omega$, let $T'(x)=\{s\in T: x\upharpoonright m(s) \text{ is a subsequence of } s\}$, where $m(s)$ is the least integer $m$ such that $x(m)$ is greater than the last (greatest) entry of $s$. Then, $T'(x)$ is a tree and $x\in A$ if and only if $T'(x)$ is ill-founded. This doesn't improve the complexity, but might suggest how to show that $A$ is proper analytic (for certain choices of $T$).