Assume that $A$ and $B$ are nonempty, bounded above, and satisfy $B\subseteq A$, show that $\text{sup}(B)\le\text{sup}(A)$
By Axiom of completeness(AoC) both A and B has supremum. Suppose $$\text{sup}(B)>\text{sup}(A)\implies \epsilon_0=\text{sup}(B)-\text{sup}(A)$$
Now $$\text{sup}(B)-\epsilon_0<b\quad\exists b\in B$$
$$\text{sup}(B)-\text{sup}(B)+\text{sup}(A)<b\quad\exists b\in B$$
$$\text{sup}(A)<b\quad\exists b\in B$$
But $\text{sup}(A)<b$ contradict the definition of sup and our assumption $B\subseteq A$ Hence $\text{sup}(B)\ngtr\text{sup}(A)\implies\text{sup}(B)\le\text{sup}(A)$
Is my proof is correct$?$
The proof is correct, but I dislike your $\implies$ sign. Instead, after having written that $\sup B>\sup A$, you could write “let $\varepsilon_0=\sup B-\sup A$”. And in the three lines after the word “now”, you could explain how to jump from each line to the next one.
Anyway, I would just say that, since $B\subset A$, each upper bound of $A$ is also an upper bound of $A$ and that therefore the least upper bound of $A$ (which is $\sup A$) must be an upper bound of $B$. Since $\sup B$ is the least upper bound of $B$, it follows that $\sup B\leqslant\sup A$.