Help with a proof of a proposition about countable sets The proposition and proof, as seen in Analysis 1 textbook, by Vladimir A. Zorich:

An infinite subset of a countable set is countable.


Proof.  It suffices to verify that every infinite subset $E$ of $\mathbb{N}$ is equipollent with $\mathbb{N}$. We construct the needed
bijective mapping $f : \mathbb{N} → E$ as follows. There is a minimal
element of $E_1 := E$, which we assign to the number $1 ∈ \mathbb{N}$
and denote $e_1 ∈ E$. The set $E$ is infinite, and therefore $E_2 :=$
$E_1 \setminus e_1 $ is nonempty. We assign the minimal element of
$E_2$ to the number 2 and call it $e_2 ∈ E_2$. We then consider $ E_3$
$:= E \setminus \{e_1,e_2\}$, and so forth. Since $E$ is an infinite
set, this construction cannot terminate at any finite step with index
$n ∈ \mathbb{N}$. As follows from the principle of induction, we
assign in this way a certain number $e_n ∈ E$ to each $n ∈$
$\mathbb{N}$. The mapping $ f : \mathbb{N} → E $ is obviously
injective.
It remains to verify that it is surjective, that is, $f (\mathbb{N}) =$
$E$. Let $e ∈ E$. The set $\{n ∈ \mathbb{N} |n ≤ e\}$ is finite, and hence the
subset of it $\{n ∈ E | n ≤ e\}$ is also finite. Let $k$ be the number
of elements in the latter set. Then by construction $e = e_k$ .
End of proof.

The part where it is proven that the function is surjective is the part I don't understand. It looks to me like the final conclusion is missing. I am looking for some insight and the idea on that part of the proof since I have no clue what is happening there.
 A: Well informally what this is doing is putting all the elements of $E$ in order and labeling the first element $e_1$ then second element $e_2$ etc.
Consider the element $e\in E$.  $e$ is a natural number.  There are $e$ elements of $\mathbb N$ that are less then or equal to $e$. Or in other words $G=\{n\in \mathbb N| n\le e\} = \{1,2,3,4,......,e\}$ a finite subset of $\mathbb N$.
Now let $F= \{n\in E| n \le e\}\subset G$ so whereas $|G| = e$ then $|F| = k\le e$.
Now consider the set $\{e_1, e_2, e_3,....,e_k\}$.
I claim the set is $F$.  If $1 < k$ then $e_1 =\min E$ and that can be $e$ because that would mean that that $F= \{n\in D| n\le e\} = \{e\}$ and that has only one element.  So $e_1 < e$ so $e_1\in F$.
And so on.  As each $e_{i<k} = \min E_i$ and there were more than $i$ elements less than $e$, $e_i \in F$.
Now finally after we do this $k-1$ times when we get to $e_k = \min E \setminus \{e_1,e_2,....,e_{k-1}\}$ and there were $k$ elements in $E$ that were less than or equal to $e$ and we've removed $k-1$ of them, then there is only one element in $E$ that is less than or equal to $e$ left.  And that element is $e$ itself!
So $\min E \setminus \{e_1,e_2,....,e_{k-1}\}= e$.
And so we assigned $e_k = e$.
So $e$ was not overlooked and this process will get $e$ eventually.  So no elements of $E$ will escape.  And these process is surjective.
A: The idea is that $e_k$ is defined to be the $k$th smallest element of $E$. The last part of the proof shows that for every $e \in E$, there is some $k$ s.t. $e$ is the $k$th smallest element of $E$. This is because there are exactly $k$ elements of $E$ which are $\leq e$ (by the definition of $k$).
A: The key in Zorich's demonstration is that he is using the natural numbers as the "larger" set (it is the image of any countable set under the appropriate bijection) and then taking E as a subset of N (since any member of the infinite subset of the countable set can thus be put in correspondence with a member of N). Rudin in my view is more explicit, PMA 3ed, theorem 2.8.
