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 $x=1$?