Here are two false proofs of the fact that $\Bbb Q$ is uncountable. From Stephen Abbott's Understanding Analysis.
Proof 01
Lemma:(Nested Interval Property – NIP). For each $n ∈ N$, assume we are given a closed interval $I_n = [a_n, b_n]$. Assume also that each I_n contains I_{n+1}. Then, the resulting nested sequence of closed intervals
$$I_1 ⊇ I_2 ⊇ I_3 ⊇ I_4 ⊇ · · ·$$
has a nonempty intersection.Assume, for contradiction, that $\Bbb Q$ is countable. Thus we can write $\Bbb Q =\{r_1, r_2, r_3, . . .\}$ and, construct a nested sequence of closed intervals with $r_n \not \in I_n$. Our construction implies $\bigcap_{n=1}^{\infty} I_n = ∅$ while NIP implies $\bigcap_{n=1}^{\infty} I_n \neq ∅$
. This contradiction implies $\Bbb Q$ must therefore be uncountable.
Proof 02
Theorem : The open interval $(0,1)$ is uncountable.
Proof: Lets assume that $(0,1)$ is countable and thus let $f:\Bbb N\to (0,1)$ be a bijective function. Then let $f(m) = 0.a_{m1}a_{m2}a_{m3} . . . .$ (decimal representation). Let $x=0.b_1b_2…$ with $b_i=3$ if $a_{ii}=2$ else $b_i=2$. Its not difficult to see x is a different number from the counted set a contradiction.
Now the question is: Every rational number has a decimal expansion, so we could apply this same argument to show that the set of rational numbers between $0$ and $1$ is uncountable. However, because we know that any subset of $\Bbb Q$ must be countable, the proof of Theorem.
I can't figure out the flaws in these two arguments.
Best Answer
In the first case, you've ignored the possibility that your sequence of intervals is, say, $I_n = [\pi - 1/n, \pi + 1/n]$. Then $\bigcap_nI_n$ is nonempty, but the only element it includes is $\pi$, which is not rational. Importantly, the NIP holds only within the reals, not the rationals.
In the second proof, observe that the number you construct (0.3233222232333... or some such) is going to be nonterminating - and it would be very easy to concoct a listing of rational numbers so that the decimal was also nonrepeating. For example, say our list was 0.2, 0.03, 0.002, 0.0002, 0.00003, 0.000003, ...; repeating in increasing "patches". Then the "diagonal" number we construct is a nonrepeating decimal - which is necessarily irrational. Thus the diagonal argument only tells us that our list of rationals is missing an irrational - which we already knew, because that was what it was supposed to do.