Every nonempty perfect subset of $\mathbb{R}$ is uncountable

Suppose, for contradiction, that the nonempty perfect set $P$ is at most countable; by [L5] fix a surjection $f:\mathbb{N}\to P$.

By [L6] fix a bijection $\beta :\mathbb{N}\to \mathbb{Q}$ and put $e:=\iota \circ \beta$ with $\iota (q)=\widehat{q}$, a bijection from $\mathbb{N}$ onto ${\mathbb{Q}}_{\mathbb{R}}$.

Recall the terminology of the Given: a pair $(p,q)$ of elements of ${\mathbb{Q}}_{\mathbb{R}}$ is good when $p<q$ and $(p,q)\cap P\mathrm{\ne }\mathrm{\varnothing }$, and $G$ is the set of good pairs.

Refinement claim. For every good $(p,q)$, every $x\in \mathbb{R}$ and every real $\eta >0$ there is a good $(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ with $[p^{\mathrm{\prime }},q^{\mathrm{\prime }}]\subseteq (p,q)$, $x\mathrm{\notin }[p^{\mathrm{\prime }},q^{\mathrm{\prime }}]$ and $q^{\mathrm{\prime }}-p^{\mathrm{\prime }}<\eta$. To see it, fix $y\in (p,q)\cap P$ and, $(p,q)$ being open, a real $\epsilon >0$ with $N_{\epsilon }(y)\subseteq (p,q)$; since $y\in P$ is not isolated, [L1] gives $z\in N_{\epsilon }^{\ast }(y)\cap P$, so $z\in P\cap (p,q)$ and $z\mathrm{\ne }y$. At least one of $y,z$ differs from $x$; let $w$ be $y$ if $y\mathrm{\ne }x$ and $z$ otherwise, so $w\in P\cap N_{\epsilon }(y)$ and $w\mathrm{\ne }x$. Put $\delta :=\min \{\epsilon -\mathrm{\mid }w-y\mathrm{\mid },\mathrm{\mid }w-x\mathrm{\mid },\eta \cdot 2^{-1}\}$, a positive real by [L11] since each entry is positive, and use [L6] to fix $p^{\mathrm{\prime }},q^{\mathrm{\prime }}\in {\mathbb{Q}}_{\mathbb{R}}$ with $w-\delta <p^{\mathrm{\prime }}<w<q^{\mathrm{\prime }}<w+\delta$. Then $[p^{\mathrm{\prime }},q^{\mathrm{\prime }}]\subseteq N_{\delta }(w)\subseteq N_{\epsilon }(y)\subseteq (p,q)$ by [L3], the pair $(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ is good because $w\in (p^{\mathrm{\prime }},q^{\mathrm{\prime }})\cap P$, the point $x$ lies outside $N_{\delta }(w)$ because $\mathrm{\mid }x-w\mathrm{\mid }\ge \delta$, and $q^{\mathrm{\prime }}-p^{\mathrm{\prime }}<2\delta \le \eta$.

Successor rule. For $(k,(p,q))\in \mathbb{N}\times G$ let $m$ be the least natural for which some natural $n$ makes $(e(m),e(n))$ good with $[e(m),e(n)]\subseteq (p,q)$, $f(k)\mathrm{\notin }[e(m),e(n)]$ and $e(n)-e(m)<1\mathrm{/}(k+1)$, and let $n$ be the least natural with those properties for that $m$; put $T(k,(p,q)):=(\sigma (k),(e(m),e(n)))$. The set of eligible $m$ is nonempty by step 2.1 applied with $x=f(k)$ and $\eta =1\mathrm{/}(k+1)>0$, since $e$ is onto ${\mathbb{Q}}_{\mathbb{R}}$, so both minima exist by [L7] and $T:\mathbb{N}\times G\to \mathbb{N}\times G$ is a total function defined without any selection.

The recursion. $P$ is nonempty, so fix $y_0\in P$ and, by [L6], elements $p_0<y_0<q_0$ of ${\mathbb{Q}}_{\mathbb{R}}$; then $(p_0,q_0)$ is good. Apply [L8] with $Y=\mathbb{N}\times G$, seed $(0,(p_0,q_0))$ and map $T$ to get $h:\mathbb{N}\to \mathbb{N}\times G$ with $h(0)=(0,(p_0,q_0))$ and $h(\sigma (k))=T(h(k))$; an induction on $k$ shows the first coordinate of $h(k)$ is $k$, so write $h(k)=(k,(p_k,q_k))$ with every $(p_k,q_k)$ good.

Writing $I_k:=[p_k,q_k]$ and ${\mathrm{\ell }}_k:=q_k-p_k>0$, the rule of step 3.1 gives, for every $k\in \mathbb{N}$: $I_{k+1}\subseteq (p_k,q_k)\subseteq I_k$, so the intervals are nested and nonempty; $f(k)\mathrm{\notin }{I}_{k+1}$; ${\mathrm{\ell }}_{k+1}<1\mathrm{/}(k+1)$; and $I_k\cap P\mathrm{\ne }\mathrm{\varnothing }$, because $(p_k,q_k)\cap P\mathrm{\ne }\mathrm{\varnothing }$.

For every real $\epsilon >0$ there is $k\ge 1$ with ${\mathrm{\ell }}_k<\epsilon$, and moreover ${\mathrm{\ell }}_k\to 0$: by step 5.1 one has ${\mathrm{\ell }}_j<1\mathrm{/}j$ for every $j\ge 1$, since ${\mathrm{\ell }}_j={\mathrm{\ell }}_{(j-1)+1}<1\mathrm{/}((j-1)+1)=1\mathrm{/}j$; given $\epsilon >0$, [L10] supplies a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$, and then every $k\ge n$ satisfies $k\ge 1$ and $\mathrm{\mid }{\mathrm{\ell }}_k-0\mathrm{\mid }={\mathrm{\ell }}_k<1\mathrm{/}k\le 1\mathrm{/}n<\epsilon$ by [L10] and [L13], which is both assertions, the second by [L12] since a rational $\epsilon >0$ is in particular a real one.

By [L9] the nested family $(I_k)$ of nonempty closed bounded intervals has an intersection that is a single point, since its lengths tend to $0$ by step 6.1; write $\{x\}$ for it, so $x\in I_k$ for every $k$.

$x\in P$: let $\epsilon >0$ be real and use step 6.1 to fix $k\ge 1$ with ${\mathrm{\ell }}_k<\epsilon$; by step 5.1 there is $y\in I_k\cap P$, and $x\in I_k$ by step 7.1, so $\mathrm{\mid }y-x\mathrm{\mid }\le {\mathrm{\ell }}_k<\epsilon$ by [L13] and $y\in N_{\epsilon }(x)\cap P$. Every neighbourhood of $x$ therefore meets $P$, so $x\in \bar{P}=P$ by [L1] and [L2].

For every $k\in \mathbb{N}$ one has $x\in I_{k+1}$ by step 7.1 while $f(k)\mathrm{\notin }{I}_{k+1}$ by step 5.1, so $x\mathrm{\ne }f(k)$; thus the element $x$ of $P$ found in step 8.1 is not a value of $f$, contradicting the surjectivity of the $f$ fixed in step 1.1. The assumption is therefore untenable: a nonempty perfect subset of $\mathbb{R}$ is not at most countable, that is, it is uncountable.