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. 

Let $C=f(A)$ the image of $f$ with the order induced by $B$. Every non-empty subset of $C$ has minimum and maximum. This implies that every element in $C$ distinct from the minimum has immediate predecessor and every element distinct from the maximum has immediate successor. Let $c_1$ the minimum of $C$. For every natural number $n$, we choose $c_{n+1}$ in $C$ as the immediate succesor of $c_n$ if $c_n$ is distinct from the maximum of $C$. If this process stop we are done. Otherwise we have a a strictly order preserving function $g:\mathbb{N}\to C$ wich is not suryective. Let $c$ be the minimum of $C\setminus g(\mathbb{N})$. And let $d$ the immediate predecessor of $c$ in $C$. By election of $c$, there exists a natural number $k$ such that $d=c_k$. Then $c=c_{k+1}$ is in $g(\mathbb{N})$. A contradiction.
EDIT:
We can also follow your approach in this way: For every natural number $n$, let $a_n$ in $A$ such that $f(a_n)=c_n=g(n)$. Let $D=(a_n)_{n\in\mathbb{N}}$ and consider the restriction $f:D\to f(D)$. Then the bijection $f^{-1}\circ g:\mathbb{N}\to D$ is a strictly order reversing function and $D$ is a well-ordered set. This is imposible by the argument above.
