The conclusion is clearly false, for example let $C$ be any finite set without a maximum element. I'm guessing that we're being asked to prove Zorn's Lemma for subsets of $\mathbb{N}$, which says that there must be a _maximal_ element. That's easy enough: We will construct a maximal set $S$ by going through the natural numbers in order, and adding $k$ to $S$ so long as $S$ will still be contained in some set in $C$. The chain condition insures that our final set $S$ will be contained in $C$. And if there is a $T \in C$ such that $S \subsetneq T$, then take $k = \min \{T \backslash S\}$. When we got to $k$ in the construction of $S$, we would have added $k$, as $S$ would still be in $T$, so $k \in S$, a contradiction. So $S$ is maximal in $C$.

Deedlit wrote: The conclusion is clearly false, for example let $C$ be any finite set without a maximum element. I'm guessing that we're being asked to prove Zorn's Lemma for subsets of $\mathbb{N}$, which says that there must be a _maximal_ element. Thanks for the correction