Real analysis - proof approach help I am taking a course in Real Analysis this semester and thought I would work ahead a little bit.
I am just reading for the moment, but came across an interesting exercise, and was wondering if I could see how one of you would solve it.
Assume that $A$ and $B$ are nonempty, bounded above, and satisfy $B⊆A$. Show $\sup B≤\sup A.$
Not so much that the question perplexes me, but rather I wanted to see an example of a rigorous proof for this question.

I have now attempted this by myself and will post my proof here for critique. I welcome the most prudent critiques you can offer.
Firstly, since $A$ and $B$ are nonempty, and bounded above, they satisfy the axiom of completeness and thus we know that $\sup A$ and $\sup B$ exist.
Since $B \subseteq A$, then for all $b \in B$ it follows that $b \in A$.
Thus $\sup A \geq b, \space \forall b \in B \subseteq A$
Secondly, if we define $z$ as any upper bound for $B$, then by definition
$\sup B \leq z$.
Since we have established that $\sup A$ is an upper bound for $B$, 
then it is clear that $\sup B \leq \sup A$
$\text{QED} ^{\tiny{brah}}$
