If $A \subset B, a_0 \in A$ and $\beta$ is an upper bound of B then $\sup A \le \beta.$
$\textbf{Proof:}$
Since $a_0 \in A,$ then $A \ne \emptyset,$ thus $\sup A$ exists. Since $A \subset B,$ then $\sup B$ is an upper bound of $A$. By definition, $\sup A$ is the least upper bound of $A.$ This tells us that $\sup A \le \sup B.$ Also, since $\beta$ is an upper bound of $B,$ then $\sup B \le \beta.$ This implies $\sup A \le \sup B \le \beta,$ hence $\sup A \le \beta.$
Please tell me if this is a correct way in proving such a statement. Any constructive criticism would be much appreciated. Thank you very much.
I agree. It seems fine. But I'm wondering if you can't just say that, if $\beta$ is an upper bound for $B$ and $A\subseteq B$, then $\beta$ is an upper bound for $A$ and hence $\sup A\leq\beta$. Maybe we have to be extra careful because of the versions of the definitions we have in mind? Saying that $A\subseteq B\Rightarrow \sup A \leq \sup B$ seems like a stronger thing to use in your reasoning than just $A\subseteq B\Rightarrow$ "an upper bound for $B$ is an upper bound for $A$" and "$\beta$ is an upper bound for $A$" $\Rightarrow \sup A \leq \beta$. Think about it and let me know what you think. :)