Proof Verification: All subsets of the natural numbers are at most countable.

Question

Proof: Suppose $X\subset\Bbb{N}$ and $X\neq\emptyset$; By the well-ordering principle $X$ has a minimum element. $\forall X'=X\setminus\{x_0\}\subset\Bbb{N}$ such that $x_0\leq m,\forall m\in X$, we know that that $X'$ still has a minimum element $x_1$, by the well-ordering principle. Clearly, $x_1$ is the next least element in $X$. By defining a function $f:\Bbb{N}\rightarrow X$ such that if $x_n$ is the least element of $X_n\subset\Bbb{N}$ and $x_{n+k}$ is the least element of $X_n=X\setminus\{x_n,x_{n+1},...,x_{n+k-1}\}\subset\Bbb{N}$, then \begin{align}f(0)=x_n\quad and\quad f(n+k)=x_{n+k}\end{align} for all $n,k\in\Bbb{N}$. Thus, $f$ is surjective. Furthermore, if $f(j)=f(i)$ then $x_{n+j}=x_{n+i}$. Since $x_{n+i}$ is the least element of $X_{n+i}=X_n\setminus\{x_n,...,x_{n+i-1}\}$ and $x_{n+j}$ is the least element of $X_{n+j}$. We know by the well-ordering principle that the least element is unique. Suppose $X_{n+i}\neq X_{n+j}$ such that $X_{n+i}\subset X_{n+j}$ where $X_{n+j}=X_n\setminus\{x_n,...,x_{n+i},...,x_{n+j-1}\}$. Since $x_{n+i}\notin X_{n+j}$, then it cannot be the least element of $X_{n+j}$ and $x_{n+i}\neq x_{n+j}$, a contradiction. Therefore, it must be that $X_{n+i}=X_{n+j}$ which means $\vert X_{n+j}\vert =\vert X_{n+i}\vert$ and $i=j$. Hence, $f$ is also injective.