I fear I over simplified the following problem:
For any partially ordered set $(A,\leq)$, let $A^* = A- \{\max A,\min A\}$ if $\max A$ and $\min A$ exist. Show the inclusion $(A^*,\leq)\hookrightarrow (A,\leq)$ is $\sup$-continuous.
So I took any set $B\subseteq A^*$ such that $\sup B$ exists. Then $\iota(B)=B$, and $\iota(\sup B)=\sup B$. Then $\iota(\sup B)=\sup B=\sup(\iota(B))$, and so $\iota$ is $\sup$-continuous.
This seems too simple so I'm sure I've misinterpreted something. Can someone point out the source of error?
The set of upper bounds depends on whether you are in A or in A*, so technically you have to check that sup(B) actually is the least upper bound of B in A. It would still be an upper bound of B, but there are 2 new contenders, max(A) and min(A). Of course sup(B) < max(A), so max(A) is a larger upper bound and presents no problem. On the other hand, min(A) < sup(B), so if min(A) is an upper bound for B in A, then it must in fact be the sup of B in A. This could only happen if B is empty, and it could happen. For example, let A={0,1,2} with the usual order and let B be the empty set. Then "sup(B)" in A*={1} is 1, but the sup of B in A is 0. This problem occurs only because A* has a least element, thus making the empty set have a sup in A*. Everything is fine if B is nonempty.
If I've correctly hit on the point you were stumbling over, perhaps it would help to adjust notation to keep in mind that sups are conditional, e.g., $\sup_A (B)$ for the sup of $B$ in $A$.