Proving that $(0,1)$ is uncountable I have this proof:
"Suppose by contradiction that $(0, 1)$ is countable. This means we can find a
sequence $(x_n)$ such that for every $x ∈ (0, 1)$ there exists a unique $n ∈ N$ with $x = x_n$.
Now we inductively construct a nested sequence of closed intervals as follows. Start
with $I_1$ a closed interval in $(0, 1)\setminus\{x_1\}$. Next, inside $I_1$ choose a closed interval $I_2$
in $(0, 1)\setminus\{x_1\}$ (if $x_2$ is not contained in $I_1$ we set $I_1\setminus\{x_2\} = I_1$).
Proceed inductively by choosing $I_{k+1}$ in $I_k\setminus\{x_{k+1}\}$. We obtain a nested sequence of closed non-empty
intervals inside $(0, 1)$. Hence, there must be a point $y$ that is in all these intervals. By
the above* $y = x_m$ for some $m$. But by construction $x_m$ is not contained in $I_m$. This
is the contradiction that we needed to conclude the proof."
*The above is a statement saying that for a sequence of nested closed non-empty intervals, the intersection of all the intervals in the sequence is non-empty.
I'm not able to follow this proof fully. What does it mean when it says "if $x_2$ is not contained in $I_1$ we set $I_1\setminus\{x_2\} = I_1$"? Isn't that already the same as $I_1$?
Also, how do we know we can always find a closed interval $I_{k+1}$ within $I_k\setminus\{x_{k+1}\}$? Is there some precise way of doing this?
 A: For your first question, I guess they were just saying that you can take $I_2=I_1$ because it already doesn't contain $x_2$.
As for your second question: if you have a non-degenerate interval $[a, b]$ ($a<b$) and a point $x\in[a,b]$, then you can always find a smaller interval contained in $[a,b]$ which does not contain $x$: for example, if $l=b-a$ is the length of the whole interval, then at least one of the intervals $\left[a, a+\frac{l}{3}\right]$ and $\left[b-\frac{l}{3}, b\right]$ doesn't contain $x$. (They are disjoint so they cannot both contain it.)
A: I'll answer both you questions at once.
You have a closed interval $I_k$ and lets assume the endpoints are $a,b$ so $I_k =[a,b]$
Our goal is to find a closed inveral $I_{k+1}$ so that $I_{k+1} \subset I_k$ and $x_{k+1}\not \in I_{k+1}$
That is to say we need to find $I_{k+1} = [c,d]$ so that:  $a \le c < d \le b$ and we do not have $c \le x_{k+1} \le d$.
This is easy enough.

*

*If it is not the case that $x_{k+1} \in a,b$ we can pick any $c,d$ so $a \le c < d \le b$.

*And if it is the case that $a\le x_{k+1} \le b$ we can pick $c,d$
that they are on one side or the other of $x_{k+1}$ but still between
$a$ and $b$.[1]

.......
How we pick these are entirely up to us.
The text seems a little concerned that if $x_{k+1} \not \in I_k$ that we will have trouble figuring $I_{k+1} \subset I_k\setminus \{x_{k+1}\}$ and tells us explicitly to let $I_{k+1} = I_k = I_k\setminus\{x_{k+1}\}$.
I'm not sure why that text thought that'd be confusing.
We can, of course, let $I_{k+1}$ be any other valid closed interval we want.
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[1]  If this is too casual we can set up an algorithm:
Let $I_k = [a_k, b_k]$ where $a_k < b_k$.

*

*If $x_{k+1}< a_k$ or $x_{k+1} > b_k$, let $a_{k+1}=a_k;b_{k+1}=b_k$ and $I_{k+1} = I_k$.


*If $x_{k+1} = a_k$ then let $a_{k+1} = a_k + \frac {b_k - a_k}2$ and $b_{k+1} = b_k$ and $I_{k+1} = [a_{k+1}, b_{k+1}]$.


*If $a < x_{k+1} \le b$ let $a_{k+1} = a_k$ and let $b_{k+1} = a_k + \frac {x_{k+1}-a_k}2$ and let $I_{k+1} = [a_{k+1}, b_{k+1}]$.
But that's very rigid method.  Any method where we simply avoid choosing $x_{k+1}$ will do.
