I'm aware that there are more general formulations of these concepts, but I'm just starting to learn these and have been looking at them in a very restricted and simple context.
Let $A \subset \omega^{\omega}$.
We say $A$ is analytic if there is $B \subseteq \omega^\omega\times \omega^\omega$ Borel such that $A = proj(B)$ ($proj$ being the projection).
We say $A$ is $\omega$-suslin if there is a tree $T \subseteq \omega^{<\omega}\times \omega^{<\omega}$ such that $A = proj([T])$ (where $[T]$ is the set of branches of the tree).
We say $A$ is $\sum^1_1(z)$ for some $z \in \omega^\omega$ if there is a computable set $P \subseteq \omega^{<\omega}\times\omega^{<\omega}\times\omega^{<\omega}$ such that $$x \in A \iff \exists y \in \omega^\omega \forall n\in \omega P(x_{\big|n}, y_{\big|n}, z_{\big|n})$$ where the bars indicate function restrictions. We say $A$ is $\sum^1_1$ if it is $\sum^1_1(z)$ for some $z \in \omega^\omega$.
Goal: Show these three notions are equivalent, i.e,
$A$ is analytic $\iff$ $A$ is $\omega$-suslin $\iff$ $A$ is $\sum^1_1$
$\omega$-suslin $\implies$ analytic is trivial.
I'm struggling with analytic $\implies$ $\omega$-suslin.My guess is to notice that the topology on $\omega^\omega \times \omega^\omega$ has a base of countable clopen sets. This implies that every Borel set is clopen (I think? that seems way too strong), and closed sets correspond to trees, which would solve the issue. However, this seems too easy; I'm pretty sure I'm missing a lot here.
As for $\omega$-suslin $\iff$ $\sum^1_1$, I'm completely lost. I'm garbage at recursion theory to begin with, and I'm not at all comfortable with the concepts being used here. I appreciate all the help I could get; I would especially appreciate being pointed to a book or a set of notes that deals with $\sum^1_1$ and the projective hierarchy more slowly and help me understand where to start at least.
I will answer the first part of the question, namely the equivalence of 1 and 2, since the second half was already answered elsewhere.
As you correctly remarked there is a correspondence between closed subsets of $\omega^\omega$ and pruned trees, giving by taking branches, so to obtain that 1 implies 2 it is enough to show that a set $A\subseteq\omega^\omega$ is analytic if and only if there is a closed $F\subseteq\omega^\omega\times\omega^\omega$ such that $A=\mathrm{proj}(F)$.
In order to do so let $A,B$ be as in 1, so that $B\subseteq\omega^\omega\times\omega^\omega$ is a Borel set such that $A=\mathrm{proj}(B)$. Let $f\colon\omega^\omega\to B$ be a continuous surjection, so that $\mathrm{proj}\circ f$ is a continuous map $\omega^\omega\to\omega^\omega$ with range $A$. Since $f$ is continuous, $\mathrm{graph}(f)\subseteq\omega^\omega\times\omega^\omega$ is closed, and $A=\mathrm{proj}(\mathrm{graph}(f))$, so that $A$ is the projection of a closed subset of $\omega^\omega\times\omega^\omega$.
The only missing ingredient is the existence of such an $f$, but this follows from the fact that every Polish space is a continuous image of $\omega^\omega$ plus the fact that if $C$ is a Borel set in the Polish space $X$, then there is a finer Polish topology on $X$ in which $C$ is clopen, hence Polish itself, and of course a continuous map remains continuous when weakening the topology of the codomain. Both of those facts are proved in Kechris or other descriptive set theory textbooks.