A set $x$ satisfies the hypothesis of Zorn's Lemma. Let $k \in x$.
$\textbf{Prove:}$
There is a $z \in x$ such that $z$ is $S$-maximal in $x$ and $z=k \vee kSz $
$\textbf{Attempt:}$
Since $x$ satisfies the hypothesis of Zorn's Lemma, we know $x\neq \emptyset$ and $x$ is partially ordered by some relation $S$ (i.e. $x$ is irreflexive and transitive under $S$) and we know every non-empty $S$-chain $y$ $\subseteq x$ has a $S$-upper bound.
So we know there exists a $z \in x$ such that $z$ is $S$-maximal in $x$.
By the definition of maximality:
$\forall a \in x$ if $zSa \implies z=a$
We were given $k \in x$. It must hold either that $k=z$ or $k\neq z$. The result follows if we know that $k=z$.
Assume $k\neq z$.
It's sufficient to show $kSz$ holds true.
This is where I become uncertain. This part of the proof will most likely relate to $S$-chains in $x$.
$\textbf{Questions:}$
For a partially ordered set in general, or one that at least satisfies Zorn's Lemma: can it be proven that the $S$-maximal element of $x$ is a member of at least one chain $y \subseteq x$ ? (In that case it will also be the $S$-upper bound for that specific chain)
This might be a spurious endeavour, but the question $\textit{seems}$ to be suggesting to me: Given some set $x$ that satisfies Zorn, if we choose an element out of the set. It is either one of the maximal elements of $x$, or it's comparable with one of the maximals of $x$.
This would be rather striking if I'm thinking about this correctly since elements of a partial order need not always be comparable.
Every singleton is a chain. So every element is contained in a chain. That much is true, and too trivial to help.
But instead you should focus your efforts on showing that for every $z\in x$, the cone above $z$ which is the set $\{z\}\cup\{y\mid zSy\}$ also satisfies Zorn's condition.