The interval $[0,1]$ is not the disjoint countable union of closed intervals. The following proof was suggested: suppose [0,1] was the disjoint countable union of closed intervals. Write the intervals as $[a_n,b_n]$. Start by showing the set of endpoints $a_n, b_n$ is closed. At first I thought this was obvious since it seems like the complement is just the union of $(a_n,b_n)$ which is open but then I thought the complement was the infinite intersection of $(0,a_n) \cup (a_n,b_n) \cup (b_n,1)$ which is not necessarily open any more.
This comes from Taylor's proof in Is $[0,1]$ a countable disjoint union of closed sets?
 A: Your first thought was correct. By hypothesis each $x\in[0,1]$ belongs to exactly one of the intervals $[a_n,b_n]$. If it isn't an endpoint of any of the intervals, it must actually belong to the open interval $(a_n,b_n)$. Thus, the complement of the set of endpoints is indeed $\bigcup_{n\in\Bbb N}(a_n,b_n)$.
A: Assume that $[0,1]=\bigsqcup_{i=1}^\infty[a_i,b_i]$. 
Start from $x_0=b_1$. Assume $b_1<1$. 
Then $x_0+\epsilon_1\in (0,1)$ lies in some $[a_i,b_i]$, where $\epsilon_1<\frac{1}{2}$. put $x_1=a_i>x_0$. Then $|x_0-x_1|<\frac{1}{2}$. Then $x_1-\epsilon_2\in (0,1)$ lies in some $[a_j,b_j]$, where $\epsilon_2<\frac{1}{4}$, put $x_2=b_j$, then 
$x_0<x_2<x_1$ and $|x_1-x_2|<\frac{1}{4}$. Continuing this process, we get
$$x_0<x_2<\ldots <x_3<x_1\qquad |x_n-x_{n+1}|<\frac{1}{2^n}$$
thus by Leibniz's criteria, $x_n\to x\in (0,1)$ with 
$$x_0<x_2<\ldots <x<\ldots <x_3<x_1$$
But $x$ cannot lie in any closed interval. For $x\in [a,b]$, if $a<x$, then it is a must that some $a<x_{2i}<x$, but $x_{2i}$ is a right point of some interval, which is contradict to the disjoint property, thus $a= x$. Similarly, $b=x$. The proof is complete. 
A: Here is an order theoretic/model theoretic proof. As far as I'm aware, it doesn't appear to be the standard approach (although there are somewhat equivalent arguments).
Suppose $[0,1]$ is the disjoint union of a family of closed intervals $\mathcal I=\{I_n\}_{n\in\omega}$. Then since $[0,1]$ inherits its linear order from $\mathbb R$, we can set $$I_a<I_b\quad\Longleftrightarrow\quad \forall x\,\big(x\in I_a\rightarrow\forall y\,(y\in I_b\rightarrow x<y)\big),$$ so that $(\mathcal I,<)$ is now a dense, countable linear order with endpoints. Hence there exists an order isomorphism $\varphi$ from $\mathcal I$ to $\mathbb Q'=\mathbb Q\cap[0,1]$. This map associates each closed interval $[p,q]$ in $\mathbb Q'$ with a closed interval $I'_{p,q}:=\bigcup_{p\leq \varphi(a)\leq q}I_a$ in $[0,1]$, where $\varphi(a)$ is a convenient shorthand for $\varphi(I_a)$.
Now take any irrational $\alpha\in [0,1]$, and let $\{p_n\}$, $\{q_n\}$ be monotone sequences of rational numbers converging to $\alpha$ from below and above, respectively. Since each $I'_{p_n,q_n}$ is compact, $\bigcap_{n\in\omega}I'_{p_n,q_n}=:I$ is non-empty. Let $x\in I$; it belongs to some $I_x\in\mathcal I$ by our hypothesis. Further let $p=\varphi(x)$, so that $p\in\bigcap_{n\in\omega}[p_n,q_n]$. Yet our choice of $\alpha$ guarantees that $\mathbb Q\cap \bigcap_{n\in\omega}[p_n,q_n]=\varnothing$, a contradiction.
This proof generalizes easily to all types of intervals in $\mathbb R$ because the theory of dense linear orders (i.e., $\mathsf {DLO}$ without restrictions on endpoints) has exactly four complete extensions, viz. $\operatorname{Th}(\mathbb Q\cap[0,1])$, $\operatorname{Th}(\mathbb Q\cap(0,1])$, $\operatorname{Th}(\mathbb Q\cap[0,1))$, and $\operatorname{Th}(\mathbb Q\cap(0,1))$; and every one of them is $\aleph_0$-categorical.
Unfortunately, even though we essentially utilized $\mathbb R$'s completeness (just like most, if not all, of the other proofs), similar reasonings do not easily apply to more general situations. Not every space has a useful order structure, and not every type of sets could be ordered analogously. So for related questions like Is $[0,1]$ a countable disjoint union of closed sets?, you need to appeal to results in analysis/topology like the Baire category theorem.
A: A full proof:
Assume that there exists a set $A$ of disjoint closed intervals that cover [0,1] and let $B$ be the set of their endpoints excluding 0,1. Since  $\bigcup_{\lambda \in A} \lambda =[0,1] \Rightarrow$ $A$ is countable $\Rightarrow$ $B$ is countable. We’ll prove on the other hand that $B$ is a perfect set, and therefore must be uncountable, thus achieving the desired contradiction. B is closed: Let $\ell$ be a limit point of $B$. $B \subset [0,1] \Rightarrow 0\leq \ell \leq 1$. $\ell \neq 1$: Let $\lambda_{1} \in A$ be the interval that contains the point 1 $\Rightarrow \lambda_{1} = [a,1] , a<1$. Since the intervals are disjoint $B\cap (a,1) = \emptyset$. By the reasoning $\ell \neq 0$. Assume that $\ell \notin B  \Rightarrow \exists \lambda \in A$ s.t. $ \lambda = [a,b], 0<a,b<1 ,\ell \in (a,b) \Rightarrow \exists e \in B $ s.t. $ |e-\ell|< \min\{\ell-a,b-\ell\} \Rightarrow e \in (a,b), e\neq a,b \Rightarrow$ the interval that one of its endpoints is $e$ intersects $\lambda \Rightarrow$ contradicition $\ell \in B $.
Every point in $B$ is a limit point of $B$:
Let $\ell \in B \Rightarrow \exists \lambda \in A$ s.t. $ \lambda = [a,\ell],  0\leq a$ or $\lambda = [\ell,b],b\leq 1$. In the first case let $\epsilon< (1-\ell) \Rightarrow [\ell,\ell + \epsilon) \in [0,1]$. Let $x\in [\ell,\ell + \epsilon)$ and let $\lambda_{x} = [a',b']$ be the interval in $A$ that contains $x$. If $a'\leq a \Rightarrow [a',a] \subseteq \lambda \cap \lambda_{x} \Rightarrow$ contradiction $\Rightarrow$ $a<a'<x \Rightarrow |a'-\ell|<\epsilon$. By similar reasoning one can show in the second case that it’s a limit point as well.
We can conclude that $B$ is perfect $\Rightarrow$ contradiction $\blacksquare$.
