let $(X, \leq_*)$ be a partially ordered set.
Assume there is an isomorphism $f: (X,\leq_*) \to (\mathbb Z, \leq)$
let $A \subseteq X$ be a well ordered subset of $X$ with an upper bound. Meaning there is an element $b \in X$ such that $\forall a \in A, a\leq_* b$.
Show that $A$ is a finite set.
I'm not sure how to approach this problem. And I'm not sure I understand what we are given too. on one hand, $(X,\leq_*)$ is partially ordered. But on the other hand, it is isomorphic to $(\mathbb Z,\leq)$ which is linearly ordered.
If $f$ is an isomorphism, you might as well assume that $X=\Bbb Z$. It's easy to prove this there, if $A$ is a well-ordered subset of $\Bbb Z$ and it has an upper bound, then it must be a subset of some interval $[-k,k]$ and so finite.
Returning to the general case, if $f$ is an isomorphism, then well-ordered sets go to well-ordered sets, and bounded sets go to bounded sets, therefore the image of $A$ under $f$ is finite, so $A$ is finite.