How can I prove $[0,1]\cap\operatorname{int}{(A^{c})} = \emptyset$? 
If $A \subset [0,1]$ is the union of open intervals  $(a_{i}, b_{i})$ such that each
  rational number in $(0, 1)$ is contained in some $(a_{i}, b_{i})$,
  show that boundary $\partial A= [0,1] - A$. (Spivak- calculus on
  manifolds)

If I prove that $[0,1]\cap\operatorname{int}{(A^{c})} = \emptyset$, the proof is complete. 
I tried to find a contradiction, but I didn't find one.
 A: The set $\partial A$ is just the closure of $A$ minus the interior of $A$. As $A$ is a union of open intervals, it's open (and thus equals its interior), and, as $A$ contains all rationals, it's dense and so its closure is $[0,1]$. 
A: Since you say you want to prove it by contradiction, here we go:
Suppose that $x\in \mathrm{int}(A^c)\cap [0,1]$. Then there is an open interval $(r,s)$ such that $x\in (r,s)\subseteq A^c$. Every open interval contains infinitely many rational numbers, so there are lots and lots of rationals in $(r,s)$. However, since $A$ contains all rationals in $(0,1)$, then the only rationals that can be in $(r,s)$ are $0$, $1$, and rationals that are either negative or greater than $1$.
Since $x\in [0,1]$, the only possibility is $x=0$ or $x=1$ (there's a small argument to be made here; I'll leave it to you). Why can we not have $x=0$? Well, if $x=0$, then $s\gt 0$, so $(r,s)$ contains $[0,\min\{s,1\})$. Are there any rationals between $0$ an d $1$ that are in $[0,\min\{s,1\})$?
And what happens if $x=1$?
A: All you need is that $A$ contains all the rationals in $[0,1]$. Now suppose, for a contradiction, that the interior of $A^c$ was non-empty -- say that it contained $x$. Then by definition $A^c$ has to contain an open interval around $x$. But every open interval contains a rational number, and by assumption there are no rationals in $A^c$.
