Prove that $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$

I want to prove $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$, where $A,B\subset \mathbb R$ are non-empty and bounded sets from above. I have reviewed similar questions and answers, but I intended to prove directly using the following definition of least upper bound:

$\sup ( A) \in \mathbb R$ and $A\subseteq \mathbb R$ if

1. $\forall a\in A$, $a \leq \sup( A)$,
2. $\forall \varepsilon >0$, $\exists b \in A$ such that $\sup( A)-\varepsilon<b$.

Here's my proof:

1. We know $A \cup B=\{x \mid x\in A \text{ or }x\in B \} $,
2. $\forall a \in A$, $a \leq \sup⁡(A) $ and $\forall b \in B$, $b \leq \sup⁡(B) $, thus $\forall a \in A$ and $\forall b \in B$, $a,b\leq \max\{\sup⁡(A),\sup⁡(B)\}$.
3. $\forall \varepsilon>0 $, $\exists k_A \in A$ s.t. $\sup( A)-\varepsilon <k_A$ and $\exists k_B \in B$ s.t. $\sup( B)-\varepsilon <k_B$. $k_A\in A\cup B$, and $\sup(A)\geq \sup(B)$ implies $\sup(A)-\varepsilon=\max\{\sup⁡(A),\sup⁡(B)\}-\varepsilon <k_A$. After considering vice versa, we can write, $\exists k\in A\cup B$, $\max\{\sup⁡(A),\sup⁡(B)\}-\varepsilon<k$.

Since both properties are satisfied, we can conclude that $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$.

Question: Is this proof correct? I have some bad feelings regarding my 3rd argument. Thank you!