Showing the existence of an infinite (strong) antichain Suppose $P$ is a poset such that there exist (strong) antichains of size $n$ for all $n \in {\bf N}$; i.e. there exist sets $S_n$ of size $n$ in $P$ such that no pair of elements of $S_n$ has a common lower bound.
Must $P$ have an infinite antichain?

If there is an infinite set of minimal elements (or rather elements that below them the order is linear), we're done.
So let's work under the assumption there are no minimal elements (or rather, every element has two incompatible smaller elements), as there are only finitely many of them, and they have to be in every maximal antichain.
Now proceed by induction: pick an element $a$, and two incompatible elements smaller than $a$, call them $a_0$ and $b_0$. Now $a'$ has two incompatible elements below it, so we can choose one to be $a_1$ and $b_1$. Proceed by induction splitting $b_n$ to $a_{n+1}$ and $b_{n+1}$. Then $\{a_n\mid n\in\Bbb N\}$ is the antichain you seek.
Choice is necessary, since without choice it is consistent there are counterexamples. For example, if $S$ is a set which is a countable union of pairs that no infinite set of pairs admits a choice function, then the tree of choice functions from finitely many pairs will satisfy this.
