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.

Thus the set $S \subseteq T$ of $s$ such that there is no splitting above $s$ is dense in $T$.  Above every $s \in S$ there is at most one branch.  If $T$ has size $\leq \omega_1$, then since every branch extends something in $S$, there are at most $\omega_1$-many branches.