Show a set is nonempty and open I've been trying to prove the following question from Abbott's Understanding Analysis, 3rd edition, Exercise 3.2.12:
Let $A$ be an uncountable set and let $B$ be the set of real numbers that divides $A$ into two uncountable sets; that is, $s \in B$ if both $\{x: x\in A \mbox{ and } x<s \}$ and $\{x: x\in A \mbox{ and } x>s \}$ are uncountable. Show $B$ is nonempty and open.
My attempt:
"$B$ is nonempty:" Suppose $B$ is empty. Then, $A$ is either a singleton or empty, and therefore A must be countable. This contradicts assumption that $A$ is uncountable, and so $B$ is nonempty.
"$B$ is open:" Note that $B$ is open $\iff$ $\mathbb{R} \setminus B$ is closed. Consider $x \in \mathbb{R} \setminus B.$ By Density of $\mathbb{Q}$ in $\mathbb{R},$ there exists a sequence $(x_n) \subseteq \mathbb{Q}$ such that $\lim x_n = x.$ Thus, $x$ is a limit point of $(x_n).$ Since $x \in \mathbb{R} \setminus B,$ $\mathbb{R} \setminus B$ is closed. Thus, $B$ is open.
I am not sure whether my attempt is correct, and would appreciate feedback/hints. 
 A: Let $x\in B$. We need to show that there exists a $\delta>0$, such that
$$
(x-\delta,x+\delta)\subset B.
$$
If not, then for every $\delta>0$, there would exists an $x_\delta\in (x-\delta,x+\delta)$, such that, one on the sets
$$
(-\infty,x_\delta)\cap A\qquad\text{and}\qquad
(x_\delta,\infty)\cap A
$$ 
is countable.
In particular, for every $n\in\mathbb N$, there would exists an $x_n\in\big(x-\frac{1}{n},x+\frac1n\big)$, such that, one on the sets
$$
(-\infty,x_n)\cap A\qquad\text{and}\qquad
(x_n,\infty)\cap A
$$ 
is countable.
In such case, one the set
$$
B_-=\{n\in\mathbb N: x_n<x\}, \qquad
B_+=\{n\in\mathbb N: x_n>x\},
$$ 
is infinite. Assume that $B_-$ is infinite, and hence we can enumerate the elements of $B_-$ as 
$$
B_-=\{k_n\}_{n\in\mathbb N}\subset \mathbb N.
$$
Then, according to our assumption, each of the sets
$$
(-\infty,x_{k_n})\cap A
$$
is countable, and so should be their union
$$
\bigcup_{n\in\mathbb N}\big((-\infty,x_{k_n})\cap A\big)=A\cap \bigcup_{n\in\mathbb N}(-\infty,x_{k_n})
$$
But $x_{k_n}\to x$, while $x_{k_n}<x$, and hence
$$
\bigcup_{n\in\mathbb N}(-\infty,x_{k_n})=(-\infty,x).
$$ 
Thus $A\cap(-\infty,x)$ is countable, which contradicts the assumption that $x\in B$. 
The case, $B_+$ countable, is treated similarly.
A: Assume $A$ is bounded, otherwise, since $A\subset \bigcup [n,n+1)$, $[k,k+1)\cap A$ is uncountable for some $k$ so we can replace $A$ with $A\cap [k,k+1)$. Let $L(s)=(-\infty, s)\cap A$, $S = \{ x : L(x) \text{ is countable }\}$. Notice that for $y\in S$, if $x<y$ then $L(x)\subset L(y)$ so $x\in S$ while $A\subset L(\sup A)$ implies $\sup A \notin S$. On the other hand, $L(\sup S)= \bigcup L(\sup S - \frac 1n)$ shows $\sup S \in S$ and hence $\sup S < \sup A$. This furnishes a decreasing sequence $a_n \in A$ converging to $\sup S$. If $B$ were empty, $U(a_n)=(a_n, \infty)\cap A$ would be countable. But then $(\sup S, \sup A)\cap A = \bigcup U(a_n)$ shows $A\subset L(\sup S)\cup (\sup S,\sup A)\cap A$ is countable: a contradiction. 

Now let $x_n \in B^c$ converge to $x$. Assume WLOG that $x_n$ is increasing. If there exists some $n$ s.t. $U(x_n)$ were countable, then $U(x)\subset U(x_n)$ would imply $U(x)$ is countable and thus $x\in B^c$. Otherwise, since $x_n \in B^c$, all $L(x_n)$ are countable, and therefore so is $L(x)=\bigcup L(x_n)$, so $x\in B^c$ and $B^c$ is closed. 
