Assume that $P=\left\langle\omega, \preceq\right\rangle$ is a partially ordered set such that for each $n \in \omega$ there are two chains $A_n$ an $B_n$ in $P$ such that $n \subset A_n \cup B_n$. Prove that there are two chains $A$ an $B$ in $P$ such that $\omega=A \cup B$.
$\star$ $\star$ Looking for advice on how to begin approaching this problem.
I am pretty sure that Zorn's lemma will play a role but I don't even know how to begin...
Here is a hint. Konig's lemma says that every finitely branching infinite tree has an infinite path. What if we make a subtree of $2^{<\omega}$ so that any path through the tree is a partition of $\omega$ into two chains of the partial order $P$?