(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.

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.
