Writing $(a,b)$ as a disjoint union of closed intervals I've been thinking about the following question:

Is it possible to write $(a,b)$ as a disjoint union of closed intervals?

My first guess was no, but then I figured the question might be hiding something subtle. I tried various things, some which I asked about on this site, and I can't seem to construct such a union. So I decided to go back to my gut feeling and prove that it cannot be done.
Attempt:
Suppose there is such a union. Then each closed interval contains a rational. Given that the sets are disjoint, any rational specifies a unique interval. Therefore the union must be a countable union, and we can list the closed intervals. Choose some listing $I_1,I_2,\dotsc$ and let:
$$(a,b)=\bigcup_{k=1}^{\infty} I_k.$$
Now define
$$J_n=(a,b)\setminus \bigcup_{k=1}^n I_k.$$
Specifically, consider that $J_2$ contains a middle interval $(a',b')$ where $a<a'<b'<b$.
Construct a sequence where each $u_n$ is arbitrarily chosen in $J_n \displaystyle\bigcap\, (a',b')$. By the Bolzano-Weierstrass theorem, $u_n$ contains a subsequence $v_n$ such that $v_n\to c$. We must have $c \in (a,b)$ so $J_{\infty}\neq \varnothing$ which is a contradiction.
Put differently, no matter how the closed intervals are chosen, we can find a sequence in $(a,b)$ that converges to a limit in $(a,b)$ which is contained in none of the closed intervals.
Is my proof correct?
 A: The general strategy can work, but you have to be more careful in choosing the sequence $v_n$. As you have it, $v_n$ might be eventually monotonic, and it might converge to an endpoint of one of the closed intervals, which would contradict your implicit claim that $\lim_n v_n\in J_\infty$. To avoid this possibility, try constructing $v_n$ in such a way that you get
$$v_0<v_2<v_4<\cdots<v_5<v_3<v_1.$$
Then, carefully prove that you indeed have $\lim_n v_n\in J_\infty$.

Edit: Here's an explicit counterexample to the original proof. It was phrased as a proof by contradiction, which is hard to work with, so I interpret the proof to claim that given a countable sequence of disjoint closed intervals in $[0,1]$, a point is found which is outside every interval.
Let:


*

*$I_1=[0.1,0.2]$

*$I_2=[0.8,0.9]$

*$I_n=[0.2+2/3^n, 0.2+3/3^n]$ for $n>2$

*$u_n=0.2+1/3^n$


Then for any subsequence $u$ of $v$, we have $\lim v = \lim u = 0.2 \in I_1$. Note that $J_\infty$ is not empty... the point is that the sequence $u_n$ doesn't help us find a member of $J_\infty$.
