Real Analysis Proof on Axiom of Completeness The Question: 
Assume $A$ and $B$ are non-empty, bounded above subset of the $\mathbb{R}$ and that $B \subseteq A$. Show that $\sup(B) \leq \sup(A)$. 
My Attempt: 
Let $A$ and $B$ be as stated. Then by the Axiom of Completeness, $\sup(A)$, and $\sup(B)$ exists and is a real number. Define $a_0 = \sup(A)$ and $b_0 = \sup(B)$. From Lemma 1.3.7., $a_0= \sup(A)$ if and only if for all $\epsilon > 0$, there exists an $a \in A$, such that $a_0 - \epsilon < a$. Similarly, $b_0 = \sup(B)$ if for all $\epsilon_{0} >0$, there exists an $b \in B$, such that $b_{0} - \epsilon_{0} < b$. 
I am trying to manipulate these inequalities in some way to get $a_{0} \leq b_{0}$. So am I even on the right track so far? Can you try to give me some hints instead of working out the problem? I would really appreciate it. Thank you very much!
Textbook: Understanding Analysis by Stephen Abbott : Exercise 1.3.4., page 18.

Hint If $B$ is a subset of $A$. Show that $\sup(A)$ is an uperbound for $B$.
Hint If $B$ is not a subset of $A$. Take $B= \{ 1,2 \}$ and $A= \{1 \}$ to get a counterexample...
