Zorn's lemma and maximal linearly ordered subsets Let $T$ be a partially ordered set. We say $T$ is a tree if $\forall t\in T$ $\{r\in T\mid r < t\}$ is linearly ordered (such orders can be considered on connected graphs without cycles, i.e. on trees). By a branch we mean a maximal linearly ordered subset in $T$.
It is easy to prove that each tree has a branch using Zorn's lemma. However, the converse is also true (I read it recently in an article). Can anybody give a sketchy proof?

Here is a nice way of proving the well-ordering principle from "Every tree has a branch":
Let $A$ be an infinite set, and let $\lambda$ be the least ordinal such that there is no injection from $\lambda$ into $A$. Consider the set $A^{<\lambda}$, that is all the functions from ordinals smaller than $\lambda$ into $A$, and order those by end-extension (which is exactly $\subseteq$ if you think about it for a moment).
It is not hard to verify that $(A^{<\lambda},\subseteq)$ is a tree. Therefore it has a branch. Take the union of that branch, and we have a "maximal" function $f$ whose domain is an ordinal $\alpha<\lambda$. Note that it has to be a maximal element, otherwise we can extend it and the branch was no branch. 
If the range of $f$ is not all $A$ then there is some $a\in A$ which witnesses that, and we can extend $f\cup\{\langle\alpha,a\rangle\}$ to a strictly larger function. Since $f$ is maximal, it follows it has to be a surjection, and therefore $A$ can be well-ordered.
(Note: If we restrict to the sub-tree of the injective functions we can pull the same trick and end up with a bijection, but a surjection from an ordinal is just as good).
From this we have that every set can be well-ordered, and therefore the axiom of choice holds.
