Show that a countable and dense set $A$ of $\mathbb{R}$ is not closed. 
Show that a countable and dense set $A$ of $\mathbb{R}$ is not closed.
Deduce that $\mathbb{R}$ is not countable.

Hint: Let $(a_n)$ be a sequence obtained by ranking the points of $A$ in a certain order, define a sequence of intervals $[b_n , c_n ]$ such that $b_{n-1} < b_n < c_n < c_{n-1}$ , whatever $n$ is, and that $[b_n,c_n]$ contains no point $a_k$ such that $k < n$.
It's hard for me to see how I can define such a sequence of intervals with only the sequence $a_n$. Then I think that I have to apply the Borel-Lebesgue theorem, saying that any compact set A is closed and bounded.
 A: Let $b_0=a_1.$  Let $c_0$ be the first $a_n$ such that $n \gt 0, a_1 \lt a_n$, and $a_0 \notin [b_0, a_n$].  This is only a potential problem if $a_1 \lt a_0$, and in that case, because $A$ is dense, we can always find some $a_n$ between $a_1$ and $a_0$.
Now proceed by induction.  Assume we have $[b_n, c_n]$ as required.  By construction, we know that $a_k \notin [b_n, c_n]$ for $0 \leq k \leq n$.  If also $a_{n+1} \notin [b_n, c_n]$, then let $b_{n+1}=b_n, c_{n+1}=c_n$.  If $a_{n+1}$ is one of the endpoints of your interval, then choose any element of $A$ in the interior of the interval to replace that endpoint.  If $a_{n+1} \in (b_n, c_n)$ (the only remaining possibility), then choose $a \in A$ such that $a \in (b_n, a_{n+1})$ and define $b_{n+1}=b_n, c_{n+1}=a$.  Again, such an $a$ always exists because $A$ is dense in $\Bbb R$.  This completes the inductive step.
You now have a nested series of compact sets.  The intersection of such a series cannot be empty.  But by construction, that intersection can't contain any element of $A$.
The intersection is in each interval, and the intersection of a set of nested intervals is either a single point $x$ (in which case the endpoints of the intervals converge to that intersection) or is itself an interval.  But the latter case can't happen because we've proved the intersection can't contain any elements of $A$, which is dense in $\Bbb R$.  Thus, the point in the intersection is the limit of the (left) endpoints of our intervals, which all are elements of $A$; in other words, $\lim b_n = x$.  Thus, we have a convergent sequence of elements of $A$ whose limit is not in $A$.  Therefore, $A$ cannot be closed.
