An infinite set has always a countably infinite subset? Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:
Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:


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*$I$ is countably infinite: That is, to say, it has the same cardinality as $\mathbb{N}$.
$$ \vert I \vert = \vert \mathbb{N} \vert $$
Then it must be that $\vert I \vert \leq \vert \mathbb{N} \vert$ and $\vert \mathbb{N} \vert \leq \vert I \vert$ i.e. there is a bijection, say $f$, such that $f: \mathbb{N} \mapsto I$.
This implies $I \subseteq \mathbb{N}$ and $\mathbb{N} \subseteq I$.
Therefore, $\mathbb{N}$ is a subset of $I$.

*$I$ is not countably infinite: 
Now, since $I \neq \emptyset$; we choose and take out some element $i_1$ from $I$.
As $I \setminus \{i_1\}$ is also not empty($\vert I \vert = 1$ otherwise), we choose and take out some $i_2$ from $I \setminus \{i_1\}$.
Similarly, continuing we end up with a countably infinite set: $$ I_c = \{i_1, i_2, i_3, \dots\}$$ from $I$ such that $\vert I_c \vert = \vert \mathbb{N} \vert$. Then it must be that $\vert I_c \vert \leq \vert \mathbb{N} \vert $ and $\vert \mathbb{N} \vert \leq \vert I_c \vert $.
Therefore, a countably infinite set is a subset of $I_c$.
All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.
 A: Being a subset cares what elements are actually in there: $A \subseteq B \iff \forall x \in A, x\in B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
A: 
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?

No. Consider the set $\Bbb R \setminus \Bbb N$ (the set of reals that are not natural) or $\Bbb R \setminus \Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $\Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $\aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
A: In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers. 
The statement "the set of natural numbers is a subset of all infinite sets"  implied that no two infinite sets are disjoint.
