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

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 $[e(m),e(n)] \subseteq (p,q)$, $f(k) \notin [e(m),e(n)]$ and $e(n) - e(m) < 1/(k+1)$, 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 $\eta = 1/(k+1) > 0$, 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 $\ell_k := q_k - p_k > 0$, 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; $f(k) \notin I_{k+1}$; $\ell_{k+1} < 1/(k+1)$; and $I_k \cap P \ne \varnothing$, because $(p_k,q_k) \cap P \ne \varnothing$.

For every real $\varepsilon > 0$ there is $k \ge 1$ with $\ell_k < \varepsilon$, and moreover $\ell_k \to 0$: by step 5.1 one has $\ell_j < 1/j$ for every $j \ge 1$, since $\ell_j = \ell_{(j-1)+1} < 1/((j-1)+1) = 1/j$; given $\varepsilon > 0$, [L10] supplies a natural $n \ge 1$ with $1/n < \varepsilon$, and then every $k \ge n$ satisfies $k \ge 1$ and $|\ell_k - 0| = \ell_k < 1/k \le 1/n < \varepsilon$ by [L10] and [L13], which is both assertions, the second by [L12] since a rational $\varepsilon > 0$ 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 $\varepsilon > 0$ be real and use step 6.1 to fix $k \ge 1$ with $\ell_k < \varepsilon$; by step 5.1 there is $y \in I_k \cap P$, and $x \in I_k$ by step 7.1, so $|y - x| \le \ell_k < \varepsilon$ by [L13] and $y \in N_\varepsilon(x) \cap P$. Every neighbourhood of $x$ therefore meets $P$, so $x \in \overline{P} = P$ by [L1] and [L2].

For every $k \in \mathbb{N}$ one has $x \in I_{k+1}$ by step 7.1 while $f(k) \notin I_{k+1}$ by step 5.1, so $x \ne f(k)$; 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.