I am almost certain that the statement in the title is True, but am not 100% sure how to prove it, or if my conclusion is valid.
My reasoning is that since $A$ and $B$ both have a maxima, then they have an upperbound which belongs to their respective sets.
As such, we can think of $A$ and $B$ as closed intervals, and their instersection should then also at least be bounded above, and closed from the right side.
Is my line of thinking correct, is there anything I have missed?
Subsets of the reals having a maximum are not necessarily closed intervals: consider for instance the set $$ B=[0,1)\cup\{2\} $$ which has $2$ as a maximum. If we take $A=[0,1]$, then also $A$ has a maximum, but the intersection $$ A\cap B=[0,1) $$ hasn't.
In more general terms, every element $x\in A\cap B$ is less than or equal to $\max A$ and $\max B$, but there's no way to find a maximum for $A\cap B$ unless the least between $\max A$ and $\max B$ belongs to the intersection or we have further information about the two sets.
For instance if we know that $A$ and $B$ are closed intervals, then the assertion is true. Suppose that $\max A\le\max B$. If $\max A\notin B$, then $\max A<b$, for every $b\in B$. But then $A\cap B=\emptyset$. Thus, if $A\cap B\ne\emptyset$, we necessarily have $\max A\in B$ and so $\max(A\cap B)=\max A$. Similarly if $\max B\le \max A$.
No. Let $A=\mathbb{Q}\cap[0,1]$, $B=(\mathbb{Q}\cap[0,1))\cup(1,2]$. Then $A\cap B=[0,1)\cap\mathbb{Q}$.
