Critique my elementary proof for a set bounded above Let $A$ and $B$ be two non-empty subsets of $\mathbb{R}$ that are both bounded above. 
$(i)$Prove that $A ∪ B$ is bounded above 
and prove 
$(ii)$ that $\sup(A ∪ B) = \max(\sup(A),\sup(B))$.

for $(i)$
1st attempt
By the axiom of completeness, $\sup A$ and $\sup B$ exist as $A$ and $B$ are both non-empty sets which are bounded above.
Then $\{A \cup B\} \leq \max(\sup A , \sup B) $ and hence $A \cup B$ are bounded above.
2nd attempt
Define $\alpha$ as some upper bound for $A$ and $\beta$ as some upper bound for $B$.
Then $\forall a \in A, \alpha \geq a \space \text{and} \space \forall b \in B, \beta \geq b$
Then $\max(\alpha,\beta) \geq \text{all} \space x \in A \cup B$ since $x \in A$ or $x \in B$ and so $A \cup B$ is bounded above.
Are either or both of these proofs rigorous?

$(ii)$
We need to show that $\sup (A \cup B) \geq \max(\sup A, \sup B)$ and also $\sup (A \cup B) \leq \max(\sup A, \sup B)$ 
For all $x \in A \cup B$, either $x \in A \implies x \leq \sup A$ or $x \in B \implies x \leq \sup B$ 
Then $x \leq \max(\sup A, \sup B)$ and $\sup (A \cup B) \leq \max( \sup A, \sup B)$
Furthermore, it's obvious that $\sup(A \cup B) \geq \sup A$ and $\sup (A \cup B) \geq \sup B \implies \sup (A \cup B) \geq \max(\sup A, \sup B)$ and thus
$\sup (A \cup B) = \max(\sup A, \sup B)$

I am still quite new to proof writing so would appreciate any and all criticisms - please be as harsh as possible.
Thanks

(i) First attempt:
In $\{A \cup B\} \leq \max(\sup A , \sup B)$, what does your notation $\{\}$ mean? If it means a set, you cannot say it is less than or equal to a number.
Second attempt is correct.
(ii) It is correct.
