Proof Explanation: Nonempty perfect sets of $(\Bbb R, |\cdot|)$ are uncountable The proof I have some trouble with:

Suppose $\varnothing\ne A\subset\Bbb R$ is perfect. $A$ is perfect, so $A$ is closed and every point in $A$ is a limit point of $A$. Hence, $A$ cannot be finite. $A$ is infinite. We want to show that $A$ is uncountable. Suppose $A$ is countable. Then, $A = \{x_1,x_2,...\}$ is an enumeration of $A$. Consider $x_1$, and an interval $[a_1,b_1]$ s.t. $x_1\in (a_1,b_1)$. Since $x_1$ is a limit point of $A$, $(a_1,b_1)\setminus\{x_1\}\cap A\ne \varnothing$ (in fact, intersects at infinitely many points). Now, choose $x_{n_1}\in A$, s.t. $x_{n_1}\ne x_1,x_2$ and consider $(a_2,b_2) \ni x_{n_1}$ s.t. $(a_2,b_2)\subset (a_1,b_1)$ and $\displaystyle b_2-a_2 \le \frac{b_1-a_1}{2}$ and $x_1,x_2\notin [a_2,b_2]$.

Q1. Why can we find such a subset? i.e. why can we find $(a_2,b_2)$ with the above properties?

Again, $(a_2,b_2)\setminus \{x_{n_1}\}\cap A$ has infinitely many points. Choose $x_{n_2} \ne x_1,x_2,x_3$. Consider $(a_3,b_3)\ni x_{n_2}$ s.t. $(a_3,b_3)\subset (a_2,b_2)$ and $\displaystyle b_3 - a_3 \le \frac{b_2-a_2}{2} \le \frac{b_1-a_1}{2^2}$.

The same question again, I do not understand the construction of these intervals!

Continuing this way, one obtains a nested sequence of closed and bounded intervals $\{[a_k,b_k]\}_{k=1}^\infty$ s.t. $\displaystyle b_k - a_k \le \frac{b_1-a_1}{2^{k-1}}$ and $x_{n_{k-1}} \ne x_1,x_2,\ldots,x_k$ with $x_{n_{k-1}}\in (a_k,b_k)$ s.t. $x_1,\ldots,x_k \notin [a_k,b_k]$.


By the Nested Interval Theorem, we can find $x\in \bigcap_{k=1}^\infty [a_k,b_k]$ s.t. $x\in A$. Observe $x\ne x_n, n\ge 1$ by the construction above. This is not possible as $A = \{x_n\}_{n=1}^\infty$. Therefore, $A$ is uncountable.

Q2. How is the last step concluded? Where is the contradiction?
It'd be very helpful if someone could explain the proof to me in detail, I am not able to follow especially the construction of the intervals $[a_k,b_k]$ and what the contradiction is. Other than that, I was able to understand several intermediate steps. Thank you!
 A: Q1. The phrasing of the proof is quite poor. You need to start by choosing $x_{n_1} \in A$ *such that $x_{n_1} \in (a_1, b_1)$* in addition to the stated requirement that $x_{n_1} \neq x_1, x_2$. Then you can first pick any interval $(a_2, b_2)$ containing $x_{n_1}$ and contained inside $(a_1, b_1)$ where the length of the interval decreases by at least a factor of 2. If you shrink $(a_2, b_2)$ to be even closer to $x_{n_1}$, you can make sure $[a_2, b_2]$ doesn't contain $x_1$ or $x_2$, so do so.
Q2. The $\cup$ should be $\cap$. Anyway, the intersection of $[a_k, b_k]$ is a single point, say $x$. Since each $(a_k, b_k)$ contains a point of $A$ (namely $x_{n_{k-1}}$), we have that $x$ is a limit point of $A$. Since $A$ is perfect, $x \in A$. Since we assumed $A$ was countable, well $x = x_m$ for some $m$. But $x \in [a_m, b_m]$ while $x_1, \ldots, x_m \not\in [a_m, b_m]$, a contradiction.
The proof could definitely be better written. The indexing scheme isn't great, it's overly formal in places while using very confusing quantifiers in others. It could use another round of polish.

Edit: A rewrite of the argument was requested; here's mine.
Lemma: Let $\varnothing \neq A \subset \mathbb{R}$ be perfect. Then $A$ is uncountable.
Proof: Suppose to the contrary that $A$ is at most countable. If $A$ were finite, then $A' = \varnothing$, a contradiction since $A=A'$ is perfect, so $A = \{x_1, x_2, \ldots\}$ is countable.
We iteratively construct a nested sequence of closed intervals $I_1 \supset I_2 \supset \cdots$ with the following properties.

*

*The midpoint of $I_k$ is in $A$.

*The length of $I_k$ is positive and less than $1/k$.

*$x_1, \ldots, x_k \not\in I_k$.

To begin the construction, pick any $x \in A$ different from $x_1$ and let $I_1$ be a closed interval centered on $x$ of positive length less than $1$ which is small enough to not contain $x_1$. Having constructed $I_1, \ldots, I_{k-1}$, pick $I_k$ as follows. Since the midpoint of $I_{k-1}$ is in $A$, which is perfect, it is a limit point of $A$. Hence there is some $y$ in the interior of $I_{k-1}$ different from $x_1, \ldots, x_k$. Let $I_k \subset I_{k-1}$ be an interval centered on $y$ of positive length less than $1/k$ which is small enough to ensure $x_1, \ldots, x_k \not\in I_k$.
By the Nested Interval Theorem and property (2), $\cap_{k=1}^\infty I_k = \{z\}$. Since the midpoint of each $I_k$ is in $A$, $z$ is a limit point of $A$, so since $A$ is perfect, $z \in A$. Hence $z=x_m$ for some $m$. But by (3), $x_1, \ldots, x_m \not\in I_m$, contradicting the fact that $x_m = z \in I_m$. $\Box$
A: Answer to Q1: Consider $x_1$ which being a limit point has the property that every open neighborhood of it will have infinitely many points of $A$.In particular, you can choose $a_1,b_1$ and denote $I_1=[a_1,b_1]$ such that $I_1\cap A$ is non empty. Note that $I_1$ contains infinitely many points of $A$ so at least one of the halves of $I_1$ will have infinitely many points of $A$. WLOG let left half of $I_1$ have infinitely many points of $A$. Let's call it $I_1'$. Choose $x_{n_1}\in I_1'-\{x_1,x_2\}$. As before, choose $a_2,b_2$ such that $x_{n_1}\in [a_2,b_2]$ and $I_2=[a_2,b_2]\subset I_1'\subset I_1$. So length of $I_2=b_2-a_2\lt (b_1-a_1)/2$. 
Answer to Q2: Note that $I_1\supset I_2\supset\cdots$. Now define $J_n= I_n\cap A$. Now verify that $J_n$'s are also nested closed and bounded sets. By nested intervals property, $\exists x\in A\cap (\bigcap_{k=1}^\infty [a_k,b_k])\implies x\in A$. $\exists m: x_m=x$. But $x_m\notin [a_m,b_m]=I_m $. Contradiction.
Alternatively, you may prove it like this also: 
Suppose on the contrary that $A$ is countable hence $A=\{x_1,x_2,... \}$. Nested closed and bounded non empty sets in $\mathbb R$ have non empty intersection. Let's make use of that property. 
Take any $r\gt 0$ and define $N_1=\{t\in\mathbb R: |t-x_1|\lt r\}$. Define $\overline N_1=\{t\in\mathbb R: |t-x_1|\le r\}$. Note that $N_1\cap P$ is non empty. 
Consider an open neighborhood $N_n$ of $x_n$. $N_n$ has infinitely many points of $A$. Define $N_{n+1}\subset N_n$ such that $x_n\notin \overline N_{n+1}$ and $N_{n+1}$ contains infinitely many points of $A$. In this way construct sequence $\{\overline N_n\}$. 
Define $S_n=\overline N_n\cap A$. Note that $S_n$ is closed and bounded. 
The way we have constructed the sequence $N_n$ implies that $x_k\in S_k \implies x_k\notin S_{k+1}\implies $ $\cap_{n=1}S_n$ is empty which is a contradiction as nested closed and bounded non-empty sets must have non empty intersection. 
