I have a question about infinite partially ordered sets. If I have an infinite partially ordered set $P$ where every chain has finite order and I take some maximal element $x_1$ (which must exists because our chains are finite) and then I take a maximal element $x_2\in P-{x_1}$ is it true that $\{x_1,x_2\}$ is an anti chain?
I don't think it is but I am not quite sure to find a counterexample. 
What I think is true, but not quite sure how to prove it.    $\downarrow$
If I take some element $x_1\in P$ and then consider its maximal chain $T_1$ and consider the maximal element $t_1\in T_1$ and then take some $x_2\in P-T_1$ and consider its maximal chain $T_2$ with maximal element $t_2$ and continue this, then $\{t_1,t_2,t_3...\}$ is an anti chain. 
Is any of this true?
 A: No, none of it is true.
If $x_1$ is a maximal element of $P$ and $x_2\in P-x_1$, is $\{x_1,x_2\}$ an antichain? No, this fails when $P$ is a $2$-element chain.
"Take some element $x_1$ and consider its maximal chain $T_1$." What do you mean by "its" maximal chain? You mean, a maximal chain $T_1$ containing $x_1$, picked at random from the many maximal chains containing $x_1$? (Fine, but since the element $x_1$ isn't mentioned again, why bother with $x_1$? Why not just choose a maximal chain $T_1$?) Then $T_1$ will have a maximal element $t_1$, since all chains in $P$ are finite; and then $t_1$ is a maximal element of $P$, since $T_1$ is a maximal chain.
Are you going to get an infinite antichain this way? Not necessarily. What if $P$ happens to have a greatest element? Whatever you take for $x_1$ and $T_1$, you're going to end up with that top element of $P$ as your $t_1$. Needless to say, that $t_1$ is not going to be part of an infinite antichain. Sorry!
It is true that every infinite partially ordered set has either an infinite chain or an infinite antichain. Do you know a theorem from analysis, that any infinite sequence of real numbers contains a monotonic subsequence? The theorem you're trying to prove is a generalization of that, so maybe the proof of that analysis theorem has some idea you can use.
