Show $[0,1] \neq \bigcup_{n=1}^\infty I_n,$ where $I_n = [a_n,b_n]$ non-empty and pairwise disjoint The task is to show that 
$$[0,1] \neq \bigcup_{n=1}^\infty I_n,$$ where $I_n = [a_n,b_n]$ is non-empty and $I_n \cap I_m = \emptyset$ for $n\neq m.$
At the risk of being marked as a duplicate, I have asked this question because I have been trying to solve it specifically using Baire Theorem.  This is something I could not find on the array of other answers to this question and I have been struggling with it.

First, I have shown that the set of endpoints, $E = \{a_n | n = 1,2,\dots\} \cup \{b_n | n = 1,2,\dots\}$ is closed and clearly countable.  I figure that I'd like to construct open dense sets such that their countable intersection consists of the points that $I_n$ is "missing" from $[0,1]$, and by Baire theorem, this set will be dense (and thus non-empty), which suffices for showing the question statement.
My intuition tells me to use the open dense sets $J_n = (0,1)\setminus(\{a_n\}\cup\{b_n\}),$ where $\bigcap_{n\in\mathbb{N}} J_n = \bigcup_{n\in\mathbb{N}} (a_n,b_n),$ and this is dense in $(0,1).$  I believe that this shows there are points "between" the $I_n$'s in [0,1] but not in any $I_n,$ but I'm not sure if this is enough to conclude that, or how to rigorously say that.  Am I on the right track?  Any help is greatly appreciated!
 A: Your idea of looking at $E$ is a good one.   Having shown it's countable and closed, now show that $E$ has no isolated points except for possibly 0 and/or 1.  Remove whichever of those points is isolated and call the resulting set $F$.  Now $F$ is still closed and countable, and has no isolated points.
Thinking about $F$ as a (complete) metric space in its own right, this implies that every singleton set $\{x\}$ is nowhere dense in $F$.  Use this to show $F$ is meager and contradict the Baire category theorem.
Indeed, this shows in general that a closed perfect subset of a complete metric space must be uncountable.
A: Maybe I'm making mistake supposing that this answer will be sufficient.
But I think the main argument is that if $I_{n} \cap I_{m}=\emptyset$, we can say that either $b_{n}<a_{m}$ or $b_{m}<a_{n}$, so we have non-empty open set between any two intervals. 
Suppose using the above we have arranged the intervals so that $0=a_{1}\leq b_{1}<a_{2}\leq b_{2}< a_{3} ...$ having $b_{\infty} = 1$.
All the open set are disjoint and nonempty.
But it seem to me, that the nonemptiness need to be shown.
A: Proof by contradiction : Suppose $I=\cup \{[a_n,b_n]:n\in N\}$ where each $[a-n,b_n]$ is a  proper subset of $I.$ WLOG let $b_1<1.$ Let $f(1)=1.$ Let $f(2)$ be the least $m$ such that $a_m>b_1.$ Now for $n\geq 2$ we make the inductive hypotheses:  
(i). The function $f(j)$ is strictly increasing for $j\in \{1,...,n\}.$
(ii). If $j<j'\leq n$ and $j,j'$ are both even then $b_{f(j')}<a_{f(j)}.$
(iii). If $j<j'\leq n$ and j,j' are both odd then $b_{f(j)}<a_{f(j')}.$ 
(iv). If $j\leq n$ and $j'\leq n$ where $j$ is even and $j'$ is odd then $b_{f(j')}<a_{f(j)}.$
Now for $n\geq 2,$ define $f(n+1)$ inductively as follows: 
(a).If $n$ is even let $x_n=b_{f(n-1)}$ and $y_n=a_{f(n)}.$
(b). If $n$ is odd let $x_n=b_{f(n)}$ and $y_n=a_{f(n-1)}.$ 
(c). Let $f(n+1)$ be the least $m>f(n)$ such that $x_n<a_m\leq b_m<y_n.$
Note:  $x_n<y_n.$ Any $[a_m,b_m]$ which intersects $(x_n,y_n)$ must be a subset of $(x_n,y_n).$ Otherwse it would intersect, but not be equal to, $[a_{f(n-1)},b_{f(n-1)}]$ or $[a_{f(n)}, b_{f(n)}].$  So $(x_n,y_n)=\cup F$ where $F=\{[a_m,b_m]:x_n<a_m\leq b_m<y_n\}.$  But $F$ cannot be finite, else the non-empty bounded open interval $(x_n,y_n)$ is closed. So $\{m: [a_m,b_m]\subset (x_n,y_n)\}$ is infinite. This justifies  the phrase "...the least $m>f(n)$ such that..." in line (c), above.
We have $$0\leq b_{f(1)}<b_{f(3)}<b_{f(5)}<...<a_{f(6)}<a_{f(4)}<a_{f(2)}\leq 1 .$$ Let $x=\sup_{n\in N}b_{f(2n-1)}$ and $y= \inf_{n\in N}a_{f(2n)}.$ We have $x\leq y$ and $[x,y]\subset I.$ Note that $$ (\bullet ) \quad \forall n\in N \;([x,y]\cap [a_{f(n)},b_{f(n)}]=\phi).$$ 
$(\bullet \bullet)$. We claim that $[x,y]\cap (\cup_{n\in N}[a_n,b_n])=\phi,$ contradicting $\cup_{n\in N}[a_n,b_n]=I.$ 
For if not, suppose $[x,y]\cap [a_k,b_k]\ne \phi.$  We have $k\ne 1=f(1).$ By the def'n of $f(2)$ we have $k>f(2).$ And $(\bullet ) \; k\ne f(n)$ for any $n.$ And $f:N\to N $ is strictly increasing. So there exists $n\geq 2$ with $f(n)<k<f(n+1).$ But then $x_n<a_k\leq b_k<y_n$ with $f(n)<k<f(n+1),$ contradicting the inductive def'n (line (c)) of $f(n+1).$ 
Remarks: 1. In the last sentence above, we have $x_n<a_k\leq b_k<y_n$ by the first part of the "Note" that follows line (c).... 2. Georg Cantor used  a similar technique  in his second proof that $I$ is uncountable.
A: Just a suggestion
We can suppose without loss of generality that $\sup I_n<\inf I_{n+1}$ for all $n$.  Suppose $[0,1]=\bigcup_{n=1}^\infty I_n$ where $I_n$ are disjoint and non empty. Let $J_n=]\sup I_n,\inf I_{n+1}[$. Since $I_n$ are compact, $d(I_{n},I_{n+1})>0$ and thus $J_n\neq \emptyset$. In particular, for all $n$, there is $\tilde I_n\supset I_n$ s.t. $\tilde I_n$ is of the form $]a,b[$ and s.t. all $\tilde I_n$ are disjoint. So, $\bigcup_{i=1}^\infty  \tilde I_n$ is a cover of $[0,1]$ with open set, and thus, by compacity, there is $n_1,...,n_k$ s.t. $$[0,1]\subset \bigcup_{i=1}^k \tilde I_{n_i}.$$
Finally, by hypothesis, $$\bigcup_{n=1}^\infty I_n=[0,1]\subset \bigcup_{i=1}^k I_{n_i},$$
which is a contradiction with $I_n\neq \emptyset$ for all $n$.
