<s>[INST] Cantor's "original" proof of $\mathbb{R}$ uncountable Found this problem in a set of notes I'm self studying from. I believe there might be an error. 
"Consider $\{x_n\}_{n∈N}$ a sequence in $[0, 1]$. Construct new sequences $a_n$ and $b_n$ as follows: $a_0$ and $b_0$ are the first two elements in the sequence ${x_n}$ such that $a_0 < b_0$. For $n ≥ 0$, take the first two unique elements of $\{x_n\}$ that lie in $(a_n, b_n)$, and call then $a_{n+1}$ and $b_{n+1}$, where $a_{n+1} < b_{n+1}$ (if there are none, simply let the sequence
stagnate at $(a_n, b_n)$). Prove that $\bigcap_{n≥0}(a_n, b_n)$ is nonempty and does not appear in ${x_n}$ (Hint: any element $x_n$ has the property that either there exists m such that $x_n ≤ a_m$ or $x_n ≥ b_m$). Conclude that it is not possible to enumerate $[0, 1]$."
Shouldn't it be : Prove that $\bigcap_{n≥0}[a_n, b_n]$ is nonempty? Because otherwise, we can consider the following sequence: 
$(\frac{1}{2} - \frac{1}{3}, \frac{1}{2} + \frac{1}{3} , \frac{1}{2} - \frac{1}{4}, \frac{1}{2} + \frac{1}{4} , \frac{1}{2} - \frac{1}{5}, \frac{1}{2} + \frac{1}{5}, \dots ,\frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n}, \dots)$
This would give us the nested open intervals $\bigcap_{n \geq 3} \left( \frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n} \right) = \emptyset$
 [/INST] Nope, it is correct as stated.  Your intersection $\bigcap_{n \geq 3} \left( \frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n} \right)$ is actually not empty, since it contains $\frac{1}{2}$.  Indeed, for any $n\geq 3$, $\frac{1}{2}-\frac{1}{n}<\frac{1}{2}<\frac{1}{2}+\frac{1}{n}$.
(That said, the proof would work just as well if you used $[a_n,b_n]$ instead of $(a_n,b_n)$, though you might have to say things a bit differently in the case that your sequence stagnates.)
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