The question is about a solution to the following exercise:
Find an infinite partial order that has no infinite antichain but is not a union of finitely many chains.
Let's take infinitely many finite antichains and make some order out of them. Let the first antichain be a single element one and each successive antichain to be bigger than the previous in size by $1$. Additionally, for every element in antichain $n$, there is an element in antichain $n + 1$ with which the two are related. So the diagram of this order should look like a tree divided into levels by antichains, like in a boolean lattice, but infinite and without the top half.
But does this order have an infinite antichain?
On one hand, each successive antichain is finite by definition, so there should not be any infinite antichains, yet their composition can be infinite. Similarly to infinite series of natural numbers where each partial sum is finite but the whole series is infinite.
But, on the other hand, we know that the size of given antichain is greater or equal to the size of the longest chain at its level. And it is a fact that infinite partial order must contain either an infinite chain or infinite antichain. So, since they are always equal, they should be both infinite.
The construction does not work in the stated level of generality. For instance, take an infinite complete binary tree. You obviously get infinite anti-chains there. Your levels are of size n rather than $2^n$, but you can use a single branch at each level to make it still fail.
There's no reason to connect the levels so arbitrarily though. Just model them off of complete bipartite graphs and it obviously works.
A more natural example is furnished by Dickson's lemma.