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.

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 $A^c$.
