(Another) Alternative Proof to Baby Rudin 2.8 Theorem: Every infinite subset of a countable subset A is countable.
(Note: below A is the subset of B)
I try to do the proofs on my own before reading Rudin's. Sometimes I fail heroically, sometimes comically. I can't see why I failed here, if I did.
Consider any $A\subset B$ where $B$ is countable. Assume for contradiction that $A$ is uncountable. Then, by definition, $\exists \alpha\in A : \nexists f(\alpha)\mapsto j\in J= \mathbb{Z}$. Because $B$ is countable, for $\forall\beta\in B, \exists j\in J$ and $\exists g:g(\beta)\mapsto j$ but this is a contradction because $\alpha\in A\subset B$.
edit: After a bunch of really unnecessary insults and vagueness, I see that the error is in my taking A from uncountable to there being no map onto Z. Thank you for everyone who (finally) helped me see that. It would have been very simple to point that out and explain it without all of the extracurriculars.
Also, I guess the re-tagging on this is OK, but it is literally in a book on analysis in a chapter called basic topology, so I'm not sure why my initial tags were wrong.
 A: It is important when starting out to work with the definitions. The first definition we need is that $B$ is countable if (and only if) there exists a 1-1 correspondence (a.k.a. bijection)
$$ f : B \to J = \mathbf{N}. $$
Then if $A \subset B$ is an infinite subset (it should be $B \subset A$ if you're following Rudin's statement but I'll keep with what you're doing) we need to show that $A$ also satisfies the definition of "countable". By definition, this means that we need to show that there exists a 1-1 correspondence
$$ g : A \to J. $$
What you've done in your proof is show that "for all $\alpha \in A$, $f(\alpha) \in J$". Look carefully at what it means to be countable. Does this statement look similar to the definition of countable? Where's the 1-1 correspondence?
I assume you mean to take $f$ as your 1-1 correspondence. You're correct that it is true that if $f : B \to J$ is 1-1 and $A \subset B$ then $f : A \to J$ is 1-1 (show this).
However, pause for a second and think about this this: is $f : A \to J$ onto? 
What you should realize is that the answer is: not unless $A = B$. For example, take $B = J$ and $f : B \to J$ to be the identity function: $f(n) = n$. Now consider the set $A = \{2,4,6,8,\dots\}$ of even numbers. The map $f : A \to J$ is not onto because there is nothing in $A$ that is mapped to $1$ (or $3$ or $5$ or ...). Rather, we need to construct a new function
$$ g : A \to J $$
such as $g(n) = n/2$. Then we have $g(2) = 1, g(4) = 2, g(6) = 3$, etc. This map $g$ is 1-1 and onto (show this).
I won't give you a proof of how to construct such a function $g$ in general (Rudin does that in his book). But I will give you the following picture. To say that a set $B$ is countable, means that we can number each element with the numbers $1,2,3,\dots$. So if $B$ is countable, we can write
$$ B = \{x_1,x_2,x_3,\dots\}. $$
Now let's say that $A$ is the set
$$ A = \{x_2,x_4,x_6,x_8,x_{10}, x_{12},\dots\}. $$
We want to construct a new numbering $y_n$ such that
$$ A = \{y_1,y_2,y_3,y_4,y_5,y_6,\dots\}. $$
For instance, let $y_1 = x_2$ and $y_2 = x_4$ and $y_3 = x_6$. Then
$$ A = \{y_1,y_2,y_3,x_8,x_{10}, x_{12},\dots\}. $$
Do you see how we have sort of "pushed" the first three indices down: $2 \to 1, 4 \to 2, 6 \to 3$? This is the general procedure.
A: If $A$ is countable then there is a bijection $i:A \to \mathbb{N}$, so 
we have $i(B) \subset i(A) = \mathbb{N}$, so to make things simpler
just take $A = \mathbb{N}$ to start with, and $B \subset \mathbb{N}$.
At this point you could note that since $B$ is infinite, there is a
map $j:\mathbb{N} \to B$ and invoke Schroeder Bernstein.
