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 $T$).



Is the set of subsequences of branches through a tree Borel?
With your assumption that the tree consists of increasing
sequences only, then the answer is yes, this is Borel. The reason
is that we can identify whether or not $x$ is a subsequence of a
branch through the tree simply by checking the arithmetic-in-$T$
condition that every finite initial segment of $x$ is a
subsequence of a node on the tree. We don't need any real
quantifiers. To see this, suppose that for every $n$, we have that
$x\upharpoonright n$ is a subsequence of some node of the tree
$t\_n\in T$. We may assume that the final node of $x\upharpoonright
n$ is the last element of $t\_n$. Let $T'$ be the tree of initial
segments of the $t\_n$'s, which is a subtree of $T$. Note that
because the sequences are all increasing, it follows that $T'$ is
finitely branching, because no initial segment of any $t\_n$ can be extended in $T'$ so as to 
skip over the next higher element of $x$. So we have essentially reduced to
the case where the tree is finitely branching. In particular, it
follows that $[T']$ is sequentially compact. For each $n$, pick
$y\_n\in[T']$ with $x\upharpoonright n\subset y\_n$. Because $T'$ is
sequentially compact, there is a convergent subsequence, and so we
may find a single $y\in [T']$ with unboundedly many
$x\upharpoonright n$ contained in $y$. Thus, $x$ is a subsequence
of $y$, which is a branch through $T'$ and hence through $T$, as
desired. So the set is arithmetically definable from the tree and
hence Borel.


If you don't insist on increasing sequences, then I believe the
set of subsequences can be complete analytic, and I can explain
this if you want.


