Suppose $T$ is a tree of height $\omega$ with $<2^\omega$-many branches. Then it must be the case that for all $t \in T$, there is $s \geq t$ such that all $x \geq s$ are comparable. Otherwise we could split into $2^\omega$-many branches.
Next take the Cantor-Bendixson derivative. Let $T_0 = T$, and for an ordinal $\alpha$, let $T_{\alpha+1} = T_\alpha$ minus the set of nodes above which $T_\alpha$ does not split. At limit $\lambda$, let $T_\lambda = \bigcap_{\alpha<\lambda} T_\alpha$. By the above paragraph, $T_{\alpha+1}$ is strictly smaller whenever $T_\alpha$ is nonempty. Thus the process converges to $T_\alpha = \emptyset$ at some $\alpha^*$, and this $\alpha^*$ must be $<|T|^+$, so $<\omega_2$ under your assumptions.
For every branch $b$ through $T$, let $\alpha_b \leq \alpha^*$ be the least ordinal $\beta$ such that $b$ is not a branch through $T_\beta$. Each $\alpha_b$ must be a successor ordinal. This means that for some node $s \in b$, there is no splitting above $s$ in $T_{\alpha_b-1}$, and $s \notin T_{\alpha_b}$. Thus $b$ is the unique branch through $T_{\alpha_b-1}$ extending $s$. If $s_b$ is the shortest node with this property, then $b \mapsto s_b$ is an injection. Thus the set of all branches has cardinality at most the size of $T$.