On the basis of @drhab and @Michael comments, I have formalized two different proofs. Please have a check if these two proofs are fine. Thank you so much!
My variant:
Suppose that $(A,\leq)$ is a partially ordered set in which each chain has an upper bound, and that $C$ is a chain in $A$. Then there exists a maximal $m$ of $A$ such that $c \leq m$ for all $c \in C$.
The proofs:
- @drhab approach
Let $T=\{D \subseteq A \mid C \subseteq D \wedge D \text{ is a chain}\}$. Then $(T,\subseteq)$ is a partially ordered set.
Assume $X$ is a chain in $T$. Let $Y=\cup_{x \in X}x$. Then $Y \in T$ and Y is an upper bound of $X$. As a result, $(T,\subseteq)$ satisfies condition of Zorn’s lemma. Hence $T$ has a maximal element $\bar{C}$.
Let $m$ be an upper bound of $\bar{C}$. Then $m$ is also an upper bound of $C$.
We now prove that $m$ is a maximal element of $A$. Assume $m < c$. Then $\bar{C} \cup \{c\} \in T$ and $\bar{C} \subsetneq \bar{C} \cup \{c\}$. This contradicts the fact that $\bar{C}$ is a maximal element of $T$. As a result, $m \nless c$ for all $c \in A$.
To sum up, $m$ is an upper bound of $C$ and a maximal element of $A$.
- @Michael approach
Let $u$ be an upper bound of $C$ and $T=\{x \in A \mid u \leq x\}$.
Assume $X$ is a chain in $T$. Then $X$ has an upper bound in $T$. As a result, $(T,\leqslant)$ satisfies condition of Zorn’s lemma. Hence $T$ has a maximal element $m$. $m \in T \implies m$ is an upper bound of $C$.
We now prove that $m$ is a maximal element of $A$. Assume $m < c$. This contradicts the fact that $m$ is a maximal element of $T$. As a result, $m \nless c$ for all $c \in A$.
To sum up, $m$ is an upper bound of $C$ and a maximal element of $A$.