Let $A,B \subset \mathbb{R}$. Show that $\sup(A \cup B) = \max\{\sup A, \sup B\}$
Solution 1:
Maybe you can reason it like this. Since $A \subset A \cup B \Rightarrow \sup(A)\leq \sup(A\cup B)$. Similarly, since $B \subset A \cup B \Rightarrow \sup(B)\leq \sup(A\cup B)$. Then $\sup(A\cup B)\geq \max\{\sup(A),\sup(B) \}$
Solution 2:
What you did seems more than fine. You may want to add that $A$ and $B$ are non-empty, though.
The reverse is quite easy: $\sup A\leq \sup(A\cup B)$ and $\sup B\leq \sup(A\cup B)$, as the supremum can never decrease if you consider a larger set!