We call a well-ordered set finite if any of it's non-empty proper subsets has the greatest element. The symbol $\subset$ here denotes a proper subset.
Let $A$ be a set not isomorphic to any proper subset $B\subset A$. Show that $A$ is isomorphic to a finite well-ordered set.
I am not sure about the induction step in my proof. Can you please check wether it is valid or not?
Proof. If $A=\varnothing$ the statement is trivial. Let $A$ be non-empty.
None of $B\subset A$ are isomorphic to $C\subset B.$ Indeed, if $f:B\rightarrow C$ is a bijection then a map $F$ defined as $$F(x)=\begin{cases}x, & x\in A\setminus B\\f(x), & x\in B
\end{cases}$$ is a bijection between $A$ and it's proper subset $(A\setminus B)\cup C.$ Since $A$ is non-empty, there exists $a\in A.$
Consider $A\setminus\{a\}.$ This is a set, smaller then $A.$ So, $A\setminus\{a\}$ is isomorphic to a finite well-ordered set $\alpha$. Let $a\notin\alpha.$
$\{a\}$ is obviously isomorphic to a finite well ordered singleton $\beta.$
Hence, $A$ is isomorphic to an ordinal sum $\alpha+\beta$ which is a finnite well-ordered set.
Best Answer
To be clear, you should say that you are proving the statement by induction on the cardinality of $A$. Your argument then works, assuming that you have already proved that cardinalities are well-ordered, so you can do induction on them. Given the nature of the question though it seems unlikely that you have that result available to you, since if you already had the full theory of ordinals and cardinals developed the result you are talking about would be rather trivial.