TITLE: Valid Proof that the Irrationals are Uncountable?
QUESTION [7 upvotes]: So I originally wanted to prove that the reals are uncountable, but the best solution I came up with was to prove the irrationals are uncountable so therefore the reals must be as well. I suppose my first question is, is this valid logic?
Next take any countable subset of $\mathbb{R} \backslash \mathbb{Q}$, put it as $S$. Since it's countable $\exists \; a : S \to \mathbb{N}$ whereas $a$ is bijective. Because it's bijective, $\exists \; b : \mathbb{N} \to S$ similarly bijective whereas $b$ is the inverse of $a$. Now let's form an irrational, put it as $s$. Start with the number defined as '$0.$'. The $n$th place after the decimal place the $n$th digit of $b_n$ with one added to it (if the $n$th digit of $b_n$ is $9$ put it to be $0$). It's clear that this number $s$ is not in the set $S$ so it follows that any countable subset of $\mathbb{R} \backslash \mathbb{Q}$ is a strict subset, so $\mathbb{R} \backslash \mathbb{Q}$ must be uncountable (otherwise it would apparently be a strict subset of itself...).
Now since $\mathbb{R} \backslash \mathbb{Q}$ is uncountable it follows that $\mathbb{R}$ is uncountable.
P.S. I tag this as topology because it's introduced in "basic topology" in Rudin
EDIT: as William and Thomas pointed out this isn't exactly valid, there is no guarantee that $s \in \mathbb{R} \backslash \mathbb{Q}$. So we can say let $S$ be a countable subset of $\mathbb{R}$ so $\exists \; a : S \to \mathbb{N}$ and similarly the inverse of $a$, $b : \mathbb{N} \to S$ and so form $s$ as the number $0.$ with the $n$th digit after the decimal point to be either the $n$th digit of $b_n$ (extending the number with as many $0$'s as needed) plus one or if that number turns out to be $10$, place $0$ there. Then it's clear that $s \not\in S$ so every countable subset is a strict subset so $\mathbb{R}$ must be uncountable. It then must follow that $\mathbb{R} \backslash \mathbb{Q}$ is uncountable since $\mathbb{Q}$ is countable, and taking a countable number of things from an uncountable number of things will clearly result in an uncountable number of things.

REPLY [3 votes]: See Theorem 2.14 of Rudin's PMA, the idea for which Rudin credits Cantor, who provided the canonical argument for the uncountability of the reals: See Cantor's Diagonalization Process.
How is your proof distinct from that proof? 
Indeed, if you thought of this on your own, KUDOS! 
See also Rudin's Theorem 2.43 (p. 41).

REPLY [3 votes]: This is essentially Cantor's diagonalization argument.
If you came up with this by yourself, that is really impressive. It is widely regarded to be one of the most beautiful proofs in mathematics.