Here's a proof: Let $A$ be a family of all finite sets of natural numbers partially ordered by set inclusion. Let $A' = \{A_k\}$ ($k \in K$) be any totally/simply ordered subset of $A$. Consider the set $B = \bigcup A_k$ (union of all sets in $A'$). Note that for every $k \in K$, $A_k \subset B$; hence $B$ is a upper bound of $A'$.
Since every totally/simply ordered subset of $A$ has an upper bound, by Zorn's Lemma, $A$ has a maximal element, i.e. a finite set which isn't a proper subset of any other finite set.
Since the statement 'proved' is obviously false, which step in the proof is incorrect?
I'm guessing $B$ isn't finite so couldn't be an upper bound for $A'$ because the upper bound needs to be finite so Zorn's lemma can't really be applied but I can't see how I could go about proving this. Maybe I'm wrong...