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?

Assume that we have a poset $(X,\leq)$ such that each linearly ordered subset of it has an upper bound. Define the following tree:  
$s\in T$ if $s\subseteq X$ and $(s,\leq)$ is a linear order and let the order relation of the tree be $s<t$ if and only if $s$ is an initial segment of $t$. It's easy to see that this is a tree. Now take a branch of the tree, $B$. If you take $(\bigcup B,\leq)$  you have a linearly ordered set. By the assumption it has an upper bound, $x$. This upper bound is a maximal element of $X$. Indeed if $x<y$ then $\bigcup B\cup\{y\}$ is above (w.r.t. the tree order) every element of $B$, contradicting the maximality of $B$.
