Let $A$ and $B$ be well-ordered sets, and suppose $f:A\to B$ is an order-reversing function. Prove that the image of $f$ is finite. Let $A$ and $B$ be well-ordered sets, and suppose $f:A\to B$ is an
order-reversing function.  Prove that the image of $f$ is finite.
I started by supposing not. Then we must have that the image of $f$, or the set $\{f(x)\in B:x\in A\}$, has infinite cardinality. If this is the case the we must have that $\vert{\{f(x)\in B:x\in A\}}\vert\geq \aleph_0$ which also means there exists a strictly order-preserving function $g:\mathbb{N}\to \{f(x)\in B:x\in A\}$. 
The contradiction I am trying to reach is that this would imply that there exists an order-reversing function from $\mathbb{N}$ to an infinite image which is a subset of a well-ordered set which can't happen but I don't know how to close the gap in the argument. 

By the givens, each non-empty subset of $f(A)$ has both a minimal and a maximal element. Conclude that $f(A)$ cannot contain a subset isomorphic to $\omega$
