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.

$\sup B$ is the smallest upper bound for $B$. This means that a strictly smaller number $\sup A$ cannot be another upper bound of $B$. Thus, $\mathbb R$ being totally ordered, there is a $b\in B$ such that $b\gt\sup A$. However, this means that, for every $a\in A$ you have $b\gt\sup A\gt a$, i.e. $b$ is an upper bound for $A$.

The condition about total ordering cannot be dropped. Imagine the set $\mathbb Z\cup\{a,b\}$ such that the relation $\le$ is given by:
$$x\le y\Longleftrightarrow\begin{cases}x\le y\text{ as integers}&x,y\in\mathbb Z\\\text{true}&y=b\\\text{true}&x=y\\\text{false}&\text{otherwise}\end{cases}$$
In other words, $b$ is the maximum of the whole set, on one hand bigger than all integers, on the other bigger than $a$, but the integers are not comparable with $a$. Now, take $A=\{a\}, B=\mathbb Z$ and you will see that $a=\sup A, b=\sup B$ and $b\gt a$ but none of the elements of $\mathbb Z$ is comparable to $a$ so cannot be an upper bound of $A$.
