maximal tree by Zorn's lemma In using the Zorn's lemma to show that every connected graph contains a spanning tree, we let $\{T_{\lambda}: \lambda \in \Lambda \}$ be a family of trees contained in $X$ which is totally ordered by inclusion. But how to show that $\bigcup\limits_{\lambda \in \Lambda} T_\lambda$ is a also a tree in $X$?

Actually, $\bigcup\limits_{\lambda \in \Lambda} T_{\lambda}$ may not be a tree. 
You have to show that every chain in your family of trees has an upper bound. So, given a chain $\{T_{\alpha}$| $\alpha \in A$}, take the union $T=\bigcup\limits_{\alpha \in A} T_{\alpha}$. It is straightforward to show that this is tree: you just have to show that, given any two vertices $v_1,v_2 \in T$, there is exactly one path between them. 
(proof by contradiction: assume two paths - both the paths will be contained in some $T_{\alpha_0}$, but $T_{\alpha_0}$ is a tree, so you get a contradiction )
Now, since you have a poset where every chain has an upper bound, Zorn's lemma can be applied. 
