TITLE: One thinking problem of uncountable set in real analysis
QUESTION [7 upvotes]: Assume: $E \subseteq R^2$ and $E$ is a uncountable set, try to prove that there exists $x_0 \in E$ s.t.$\forall B(x_0)$($B(x_0)$ is an open circular neighborhood, $x_0 \in B(x_0)$ and the center is unnecessarily $x_0$), $B(x_0) \cap E$ is a uncountable set.
The answer proves it reversely: assume that: $ \forall x \in E$, there exists one circular neighborhood $B(x)$($x \in B(x)$) s.t. $ E \cap B(x)$ is a countable set. Then $\forall x\in E$, choose a $r_x  \in \mathbb Q$, s.t.$x \in B(x,r_x)$. Therefore, $E=E \cap \bigcup _{x}{B(x,r_x)}=\bigcup_{r_x}{ E \cap B(x,r_x)}$,thus $E$ is a countable set, which leads to contradiction.
Q:For two different $x$, there may be one same $r_x$, which means it is not a single-shot mapping from the set of $B(x,r_x)$ to $ \mathbb Q$. So it can’t prove that the set of all $B(x,r_x)$ is countable? The answer is wrong?

REPLY [2 votes]: I thought it for a long time and I got another answer, please help me verify the answer, thanks!
Prove it by contradiction: $ \forall x \in E$, there exists one circular neighborhood $B(x)$($x \in B(x)$) s.t. $ E \cap B(x)$ is a countable set. Then $\forall x\in E$, choose a rational dot $q=(r_x,r_y)$ as the center and a rational number $ \delta$ as the radius, s.t.$x \in B(q,\delta) \subseteq B(x)$. Set $\Gamma=\{B(q,\delta)|\exists x\in E$,s.t.$x\in B(q,\delta) \subseteq B(x)$}. We claim that $ \Gamma$ is countable cause there is an injective mapping from $\Gamma$ to the subset of $\mathbb Q^3$. Thus, $E \subseteq \bigcup_{S \in \Gamma}{S} \subseteq \bigcup_{S\in \Gamma}{B(x)}$. Therefore, $E=E\cap\bigcup_{S\in\Gamma}{S}=E \cap \bigcup_{S \in  \Gamma}{B(x)}=\bigcup_{S \in \Gamma}{E \cap B(x)}$, which leads to the contradiction.