Every nonempty perfect subset of $\mathbb{R}$ has the cardinality of the continuum

Splitting claim. Let $J=(a,b)$ be a nonempty open interval with $J\cap P\mathrm{\ne }\mathrm{\varnothing }$, and let $\eta >0$ be real. Choose $x\in J\cap P$. Since $x$ is not isolated in $P$, there is $y\in P\cap J$ with $y\mathrm{\ne }x$; after swapping if necessary, take $x<y$. Put $\delta :=\min \{\eta \mathrm{/}4,(y-x)\mathrm{/}4,(x-a)\mathrm{/}2,(b-y)\mathrm{/}2\}$, a positive real. By [L1] choose rationals $r_0<s_0<r_1<s_1$ with $x\in (r_0,s_0)$, $y\in (r_1,s_1)$, each interval contained in $(x-\delta ,x+\delta )$ or $(y-\delta ,y+\delta )$ respectively. Then $J_0:=(r_0,s_0)$ and $J_1:=(r_1,s_1)$ are disjoint nonempty open intervals, their closures lie inside $J$, each meets $P$, and each has length $<\eta$.

Fix an enumeration of the rational quadruples using [L3]. By recursion on $n\in \mathbb{N}$, construct for every binary word $s:n\to \{0,1\}$ a nonempty open interval $J_s$ with rational endpoints such that: $J_s\cap P\mathrm{\ne }\mathrm{\varnothing }$; if $t$ extends $s$ then $\overline{J_t}\subseteq J_s$; sibling closures are disjoint; and every $J_s$ at level $n$ has length $<1\mathrm{/}(n+1)$. At level $0$, take the first rational interval in the enumeration that meets $P$ and has length $<1$. At the successor stage, for each of the finitely many parent words in lexicographic order, take the first rational quadruple in the enumeration that gives the two children supplied by step 1.1 with $\eta =1\mathrm{/}(n+2)$. The “first” rule makes the successor operation a function, so recursion produces one coherent family through all levels without any choice principle.

Let $\alpha :\mathbb{N}\to \{0,1\}$ be a binary sequence, and for each $n$ let $I_n:=\overline{J_{\alpha \mathrm{\mid }n}}$, where $\alpha \mathrm{\mid }n$ is the restriction of $\alpha$ to $n$. By step 2.1 the intervals $I_n$ are nonempty, closed and bounded, nested, and have lengths $<1\mathrm{/}(n+1)$; [L5] therefore makes their lengths tend to $0$, so [L4] gives a unique point $x_{\alpha }\in {\bigcap }_n{I}_n$. Each $I_n$ meets $P$ and $P$ is closed, so $x_{\alpha }\in P$. If $\alpha \mathrm{\ne }\beta$, let $n$ be the first index at which they differ; then $I_{n+1}$ for $\alpha$ and $\beta$ are closures of disjoint siblings from step 2.1, so $x_{\alpha }\mathrm{\ne }{x}_{\beta }$. Thus $\alpha \mapsto x_{\alpha }$ is an injection from $\{0,1{\}}^{\mathbb{N}}$ into $P$.

Fix a bijection $e:\mathbb{N}\to \mathbb{Q}$ from [L2]. For each real $x$ define $S_x:=\{n\in \mathbb{N}:\widehat{e(n)}<x\}\mathrm{.}$ If $x<y$, [L1] gives a rational $q$ with $x<\widehat{q}<y$; writing $q=e(n)$ yields $n\in S_y\setminus S_x$, so $x\mapsto S_x$ is an injection $\mathbb{R}\to \mathcal{P}(\mathbb{N})$. Characteristic functions identify $\mathcal{P}(\mathbb{N})$ with $\{0,1{\}}^{\mathbb{N}}$, and step 3.1 gives an injection $\{0,1{\}}^{\mathbb{N}}\to P$; composing these two injections yields an injection $\mathbb{R}\to P$. The inclusion $P\hookrightarrow \mathbb{R}$ is also injective, so [L6] gives $P\approx \mathbb{R}$.