Theorem For every set $A$, the following conditions are equivalent (assuming AC$_{\omega}$ only for $\text{(a)}\Longrightarrow\text{(b)}$):
$\text{(a)}\;A$ is an infinite set
$\text{(b)}\;A$ has a countably infinite subset
$\text{(c)}\;A$ has a proper subset to which it is equinumerous
We will prove, first, the chain of implications $\text{(a)}\Longrightarrow\text{(b)}\Longrightarrow\text{(c)}$, to then prove the chain $\text{(c)}\Longrightarrow\text{(b)}\Longrightarrow\text{(a)}$.
$\text{(a)}\Longrightarrow\text{(b)}$: Let $A$ be an infinite set. For each $n\in\omega$, let $A_n$ be the set of injective functions from $n$ to $A$. For each $n\in\omega$, the set $A_n$ is not empty, since $A$ is an infinite set. By AC$_{\omega}$, for the countable set $\{A_n|\;n\in\omega\}$, the cartesian product $\prod_{n\in\omega}A_n$ is not empty, and there exists a function $h$ of domain $\omega$ such that for each $n\in\omega$, $h(n)=h_n$ is an injective function from $n$ to $A$.
Then, we can form a $\omega$-sequence of distinct elements of $A$, first putting the value of $h_1$, then the values of $h_2$, etc.:
$$h_1(0);h_2(0),h_2(1);h_3(0),h_3(1),h_3(2);h_4(0),h_4(1),h_4(2),h_4(3);\dots$$
And then deleting each occurence of any element $a$ of $A$ after appearing for the first time. This sequence is actually infinite, because for any $n\in\omega,\;h_n$ has $n$ distinct terms, and we have deleted at most $n-1$ of them, if $n\geq 1$.
The set of all the terms of this sequence is a countably infinite subset of $A$.
$\text{(b)}\Longrightarrow\text{(c)}$: Let $C$ be a countably infinite subset of $A$. Let $f$ be a bijective function from $\omega$ onto $C$. Consider the function $g:A\longrightarrow A$ defined by: for each $a\in A$,
$$g(a)=\begin{cases}
f(n+1)\quad\text{if }a\in C\text{ and }a=f(n)\text{ for some }n\in\omega\\
a\qquad\qquad\,\text{if }a\in A\setminus C
\end{cases}$$
The function $g$ s injective from $A$ to $A$, so $A\approx\text{Im}(g)$, and since $f(0)\not\in\text{Im}(g),\;\text{Im}(g)$ is a proper subset of $A$ to which it is equinumerous
$\text{(c)}\Longrightarrow\text{(b)}$: Let $B$ a proper subset of $A$ such that $A\approx B$, and let $f$ be a bijective function from $A$ ontro $B$. Since $B$ is proper, let $a_0$ be an element of $A\setminus B$. Let $g:\omega\longrightarrow A$ be the sequence defined by
$$\begin{cases}
g(0)=a_0\\
g(n+1)=f(g(n))
\end{cases}$$
We will prove that $g$ is injective, and therefore $\text{Im}(g)$ will be a countably infinite subset of $A$.
Suppose that $g$ is not injective. Then there exists two natural numbers $m,n$ with $m<n$ wuch that $g(m)=g(n)$. Therefore, the set
$\mathcal{N}=\{n\in\omega|\;\text{ there exists }m>n\text{ such that }g(m)=g(n)\}$
is a nonempty subset of $\omega$. Let $n_0$ be its first element.
- In first place, $n_0\not=0$, because $g(0)=a_0,\;a_0\not\in B$, and if $n>0,\;g(n)\in\text{Im}(f)=B$
- Since $n_0>0$, we have that if $m\in\mathcal{N}$ is such that $m>n_0$ and $g(n_0)=g(m)$, then
$$g(n_0)=f(g(n_0-1))=f(g(m-1))=g(m)$$
And, since $f$ is injective
$$g(n_0-1)=g(m-1)$$
Which contradicts the choice of $n_0$ as the first element of $\mathcal{N}$
$\text{(b)}\Longrightarrow\text{(a)}:$ Suppose that $A$ has a countably infinite subset $B$, and let $f:\omega\longrightarrow B$ be a bijection. If $A$ was finite, we would have that $A\approx n$ for some $n\in\omega$, for some bijective function $g:A\longrightarrow n$. The composition $g\circ f:\omega\rightarrowtail n$ would be injective, and its restriction to $n+1$ an injective function from $n+1$ to $n$, in contrary to the pidgeonhole principle.
