Proof outline of a construction of injective function Let $A$ be nonempty s.t. there exists $f:A\to A$ injective but not subjective. Show there is a $g:\mathbb{N} \to A$ that is injective.
Below is my proof, is this a correct approach? Anything I'm missing?

 A: No, it is not correct. Why do you say that if $a_3=a_1$ then $f$ is surjective? If, say $f\colon\Bbb Z\longrightarrow\Bbb Z$ is defined by$$f(n)=\begin{cases}-n&\text{ if }n\in\{-1,0,1\}\\n+1&\text{ if }n>1\\-n-1&\text{ if }n<-1,\end{cases}$$then $f$ is injective and not surjective ($\pm2\notin f(\Bbb Z)$). But, if you define $(a_n)_{n\in\Bbb N}$ as you did with $a_1=1$, then $a_3=a_1$.
You can do it as follows. Let $A_0=A$ and let $A_1=f(A_0)=f(A)$. Since $f$ is not surjective, $A_1\ne A_0$. So, $A_1\varsubsetneq A_0$. Now, let $A_2=f(A_1)$. Since $A_1\subset A_0$,$$A_2=f(A_1)\subset f(A_0)=A_1.$$In fact, $A_2\varsubsetneq A_1$. In order to see why, take $y\in A_0\setminus A_1$; clearly, $f(y)\in A_1$, and I will prove that $f(y)\notin A_2$. If $f(y)\in A_2$, then $f(y)=f\bigl(f(x)\bigr)$, for some $x\in A$. But, since $f$ is injective,$$f(y)=f\bigl(f(x)\bigr)\implies y=f(x),$$which is impossible, since $y\notin A_1=f(A)$. By the same argument, if $A_3=f(A_2)$, then $A_3\varsubsetneq A_2$ and so on. Now, for any $n\in\Bbb N$, let $g(n)$ be an element of $A_{n-1}\setminus A_n$ and then $g$ is an injective map from $\Bbb N$ into $A$.
