Proof that $G_\delta$ sets in $\mathbb{R}$ are either countable,finite, or have cardinality of the continuum. Let $U$ be a $G_\delta$ set. We assume that it has empty interior and it's not countable. We have that $U=\bigcap_{i\in\mathbb{N}} A_i$ where $A_i$ are open. Wlog we can assume that $A_{i+1}\subseteq A_i$. Every open set in $\mathbb{R}$ is the countable disjoint union of open intervals. Set $A_i=\bigsqcup I^i_{a}$. Consider the tree $T=\{ I^i_a:i,a\in \mathbb{N}\}$ ordered by $\supseteq$. Given any maximal branch $\{I_{a_i}^i\}_{i\in \mathbb{N}}$ it's intersection is either empty or contains a single point. We can prune $T$ without affecting $U$. There is a bijection between the elements of $U$ and the branches of the now pruned $T$ where $x\in T$ is sent to $\{I_{a_i}^i\}_{i\in \mathbb{N}}$ where for each $i\in \mathbb{N}$ $I_{a_i}^i$ is the interval of $A_i$ that contains $x$ (the inverse of this map is the intersection of a given branch). We search for an appropriate subtree.
For each $i\in\mathbb{N}$ there exists an $I^i_a$ such that the number of branches that contain $I^i_a$ is uncountable since at each $i$ there can be at most a countable amount of $I^i_a$. Given a node $I^i_a$ that is contained in uncountably many branches there exist $I^j_x,I^j_y\subseteq I^i_a$ that are disjoint and are contained in uncountably many branches (as a short hand we say $I^j_x,I^j_y$ "split" the branches of $I^i_a$). We define a subtree where at the first step $J^0_0$ is any node that is contained in uncountably many branches and we set $S_0=\{J^0_0\}$. At the $n+1$-th step for all $J^n_x$ in $S^n$ we find any two $J^{n+1}_{x^\frown 0}$ $J^{n+1}_{x^\frown 1}$ that "split" the branches passing by $J^n_x$ (the bottom index will be a sequence in $\{0,1\}$). This subtree will have cardinality of the continuum.
Is this a correct proof? And would there be a more elegant way in constructing the subtree?
Edit: Looking back I realized that the "pruning" I did at the beginning was not as trivial as I thought. Not only do trivial branches have to be removed (i.e. branches with finite height) but there are also branches of $\omega$ height whose intersection is empty and at every level the node (seen as an open set) has uncountable intersection with $U$. To remove such branches we need to assure that all the child nodes closure is contained in it's predecessor. To this for each $I_a^i$ can be restricted such that it's intersection with $U$ is still uncountable. This can easily be done since the real line is separable. This assures that all branches have union that's a point.
 A: $\newcommand{\cl}{\operatorname{cl}}$ The basic idea of getting a subset of $U$ indexed by the branches of the complete binary tree of height $\omega$ can be carried out with a bit less machinery.
Let $\mathscr{B}$ be a countable base for $\Bbb R$, and let $V=\bigcup\{B\in\mathscr{B}:|B\cap U|\le\omega\}$; clearly $V\cap U$ is countable, so $U_0=U\setminus V$ is uncountable. Moreover, $G\cap U_0$ is either empty or uncountable for each open $G$ in $\Bbb R$.
Let $\Sigma$ be the set of finite sequences of $0$’s and $1$’s, and for each $n\in\Bbb N$ let $\Sigma_n=\{\sigma\in\Sigma:|\sigma|=n\}$. If $\sigma\in\Sigma$ and $i\in\{0,1\}$, denote by $\sigma^{\frown}i$ the sequence obtained by appending $i$ to $\sigma$; thus, if $\sigma\in\Sigma_n$, then $\sigma^{\frown}i\in\Sigma_{n+1}$. For $\sigma,\tau\in\Sigma$ write $\sigma\prec\tau$ iff $\sigma$ is a proper initial segment of $\tau$. I’ll construct open sets $G_\sigma$ for $\sigma\in\Sigma$ in such a way that:

*

*$G_\sigma\subseteq A_{|\sigma|}$ for each $\sigma\in\Sigma$,

*$G_\sigma\cap U_0\ne\varnothing$ for each $\sigma\in\Sigma$,

*$\{G_\sigma:\sigma\in\Sigma_n\}$ is pairwise disjoint for each $n\in\Bbb N$,

*$\cl G_\tau\subseteq G_\sigma$ whenever $\sigma\prec\tau$, and

*$\operatorname{diam}G_\sigma\le 2^{-n}$ whenever $\sigma\in\Sigma_n$.

Suppose that this has been done. Let $\Sigma_\omega$ be the set of infinite sequences of $0$’s and $1$’s, and note that $|\Sigma_\omega|=2^\omega$. For each $\sigma\in\Sigma_\omega$ and $n\in\Bbb N$ let $\sigma^n\in\Sigma_n$ be the initial segment of $\sigma$ of length $n$, and let $$F_\sigma=\bigcap_{n\in\Bbb N}G_{\sigma^n}=\bigcap_{n\in\Bbb N}\cl G_{\sigma^n}\;.$$
Condition (1) ensures that $F_\sigma\subseteq U$, conditions (2), (4), and (5) and the completeness of $\Bbb R$ ensure that $F_\sigma=\{x_\sigma\}$ for some $x_\sigma\in\Bbb R$, and condition (3) ensures that if $\tau\in\Sigma_\omega$ is distinct from $\sigma$, then $F_\tau\cap F_\sigma=\varnothing$ and hence $x_\tau\ne x_\sigma$. Thus, $\{x_\sigma:\sigma\in\Sigma_\omega\}$ is a subset of $U$ of cardinality $2^\omega$.
To construct the sets $G_\sigma$, let $y_{\langle\rangle}\in U_0$ be arbitrary, let $G_{\langle\rangle}=B(y_0,1)\cap A_0$, and note that $G_{\langle\rangle}\cap U_0$ is infinite (and in fact uncountable). (Here ${\langle\rangle}$ is the sequence of length $0$.) Given $G_\sigma$ for some $\sigma\in\Sigma_n$, let $y_{\sigma^{\frown}0}$ and $y_{\sigma^{\frown}1}$ be distinct points of $G_\sigma\cap U_0$. There is a positive $r<\min\left\{2^{-(n+1)},\frac12|y_{\sigma^{\frown}0}-y_{\sigma^{\frown}1}|\right\}$ such that $\cl\big(B(y_{\sigma^{\frown}0},r)\cup B(y_{\sigma^{\frown}1},r)\big)\subseteq G_\sigma$, and we set $G_{\sigma^{\frown}i}=B(y_{\sigma^{\frown}i},r)\cap A_{n+1}$ for $i\in\{0,1\}$.
Clearly $G_{\sigma^{\frown}0}\cap G_{\sigma^{\frown}1}=\varnothing$, and for $i\in\{0,1\}$ we have $G_{\sigma^{\frown}i}\subseteq A_{|\sigma^{\frown}i|}$, $G_{\sigma^{\frown}i}\cap U_0\ne\varnothing$, $\cl G_{\sigma^{\frown}i}\subseteq G_\sigma$, and $\operatorname{diam}G_{\sigma^{\frown}i}\le 2^{-(n+1)}$, so the recursive construction goes through as desired.
Of course any complete metric space can be substituted for $\Bbb R$.
