Proof of Distinction Between Finite and Infinite Sets Using Subsets

Proof: (related to "Distinction Between Finite and Infinite Sets Using Subsets")

Let $X$ be finite.
By subsets of finite sets, $S$ is finite with $|S| \le |X|.$
Since $S$ is a proper subset, we have even $|S| \neq |X|.$
Therefore, $|S| < |X|.$
By characterization of finite sets (no. $3$), there is no injective function $f:X\to S.$

Assume, $X$ is infinite.
Choose $x_1\in X,$ $x_2\in X\setminus\{x_1\},$ $x_3\in X\setminus\{x_1,x_2\},$ etc.
Since $X$ is infinite, we can continue this process for all indices $i\in\mathbb N$ be mapping $f:\mathbb N\to X,$ $f(i):=x_i.$
By construction, this is an injective function $f:\mathbb N\to X.$
Now, define the function $g:X\to S$ ($S$ being a proper subset of $X$) as follows:
$$g(x):=\begin{cases}x&\text{if }x\in X\setminus \{x_i\mid i\in\mathbb N\}\\x_{i+1}&\text{if }x=x_i\end{cases}$$
By construction, the function $g$ is injective as well.

By contraposition to finite case $”\Rightarrow”,$ if there is an injective function $f:X\to S,$ then $X$ is not finite, therefore $X$ is infinite.

By contraposition to the infinite case $”\Rightarrow”,$ if there is no injective function $g:X\to S,$ then $X$ is not infinite, therefore $X$ is finite.

q.e.d