Is there a less tedious means of achieving this proof? The next section of the problem asks to generalize this result to four sets, hence why I ask.
Given $A$, $B$, and $C$, there is a set $P$ such that $x\in P$ if and only if $x=A$ or $x=B$ or $x=C$.
Applying the axiom of pair to $A$ and $B$, there exists some set, $D$, such that $x\in D$ if and only if $x=A$ or $x=B$; more simply, $D=\{A,B\}$. Reapplying the axiom of pair to $C$ itself, there exists some set, $E$, such that $x\in E$ if and only if $x=C$ [or $x=C$]; more simply, $E=\{C\}$. Using the axiom of pair one last time, applied to $D$ and $E$, there exists some set, $H$, such that $x\in H$ if and only if $x=D$ or $x=E$; more simply, $H=\{D,E\}$. Finally, apply the axiom of union to $H$, which yields the desired set, $P$, where $x\in P$ if and only if $x\in S$ for some $S\in H$. By the definitions of $H$, $D$, and $E$, if $S\in H$, then $S=\{A,B\}$ or $S=\{C\}$. Suppose $S=\{A,B\}$. If $x\in S$, then $x=A$ or $x=B$. Now suppose $S=\{C\}$. If $x\in S$, then $x=C$. Hence, either way, ($x=A$ or $x=B$) or $x=C$. Since logical disjunction is associative, we may write $x=A$ or $x=B$ or $x=C$. In summary, to conclude: there exists some set, $P$, such that $x\in P$ if and only if $x=A$ or $x=B$ or $x=C$.
Not sure if this seems simpler or not, but I would probably use the axiom of pair to prove that $\{A,B\}$ and $\{A,C\}$ are sets, and then that $\{\{A,B\},\{A,C\}\}$ is a set; and finally apply the axiom of union to prove that $\{A,B,C\}$ is a set. This works for four elements with minimal changes.