Infinite Set is Equivalent to Proper Subset/Proof 1

Proof
Let $T$ be an infinite set.

By Infinite Set has Countably Infinite Subset, it is possible to construct a countably infinite subset of $T$.

Let $S = \left\{{a_1, a_2, a_3, \ldots}\right\}$ be such a countably infinite subset of $T$.

Create a Partition of $S$ into:
 * $S_1 = \left\{{a_1, a_3, a_5, \ldots}\right\}, S_2 = \left\{{a_2, a_4, a_6, \ldots}\right\}$

Let a bijection be established between $S$ and $S_1$, by letting $a_n \leftrightarrow a_{2n-1}$.

This is extended to a bijection between $S \cup \left({T \setminus S}\right) = T$ and $S_1 \cup \left({T \setminus S}\right) = T \setminus S_2$ by assigning each element in $T \setminus S$ to itself.

So a bijection has been demonstrated between $T$ and one of its proper subsets $T \setminus S_2$.

That is, if $T$ is infinite, it is equivalent to one of its proper subsets.

Now, let $T_0 \subsetneq T$ be a proper subset of $T$, and $f: T \to T_0$ be a bijection.

It follows from No Bijection between Finite Set and Proper Subset that $T$ must be infinite.