Set is Infinite iff exist Subsets of all Finite Cardinalities

Theorem
A set $S$ is infinite iff for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$.

Necessary Condition
We use mathematical induction on $n$.

The empty set is a subset of $S$ whose cardinality is $0$, so the base case $n = 0$ is proved.

Using the induction hypothesis, let $T$ be a subset of $S$ whose cardinality is $n$.

Let $f : \N_n \to T$ be a bijection, where $\N_n$ denotes the subset of natural numbers $\left\{{0, 1, 2, \ldots, n - 1}\right\}$.

Since $S$ is infinite, $T$ is a proper subset of $S$.

So let $x \in S \setminus T$, where $\setminus$ denotes set difference.

Then $T \cup \left\{{x}\right\}$ is a subset of $S$ whose cardinality is $n + 1$, completing the induction step.

Sufficient Condition
Assume that $S$ is finite.

Let $T$ be a subset of $S$ whose cardinality is $\left\vert{S}\right\vert + 1$, where $\left\vert{S}\right\vert$ denotes the cardinality of $S$.

Then $\left\vert{S}\right\vert \ge \left\vert{T}\right\vert = \left\vert{S}\right\vert + 1$, which is impossible.