Proving that $(0,1)$ is uncountable

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 $ale x_{k+1} le b$ 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 $I_{k+1} subset I_ksetminus {x_{k+1}}$ and tells us explicitly to let $I_{k+1} = I_k = I_ksetminus{x_{k+1}}$.

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 $a_k < b_k$.

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

2. 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 $I_{k+1} = [a_{k+1}, b_{k+1}]$.

3. 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.

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 $xin[a,b]$, 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 $left[b-frac{l}{3}, bright]$ doesn't contain $x$. (They are disjoint so they cannot both contain it.)