Given two bounded sets $A,B$ and $\sup A<\sup B$, is there an element in $B$ that works as an upper bound for $A$? Originally I worked this questions out by just saying that $m=\sup B$ and $m>\sup A$. Then drawing the conclusion that $m$ is an upper bound for $A$. I figured this was wrong because we're unsure that $\sup B$ is in $B$ at all.
I decided to approach it like this:
Let $\sup B=m$, which may or may not be in $B$. But, $m-1$ is in $B$ and $m-1\geq \sup A$.
Therefore, $m-1$ is an upper bound for $A$.
My main question arises when subtracting $1$ from $m$. Is this a logical justification? I can't think of any other way to justify it unless I know that the set $B$ has maximum, which I don't.
Thank you.

No, it's not justified, unless $A$ and $B$ are sets of integers. What you can say,is that $m -\frac12(m-\sup A)>\sup A$, and there's necessarily an element $b\in B$ such that
$$m\ge b>m-\tfrac12(m-\sup A)>\sup A.$$
since $m=\sup B$. The conclusion follows.
