Showing there exists no subintervals without digit. Let $A$ be the subset of $(0,1)$ containing the real numbers which have a 3 in their decimal expansion. 
As part of a larger result in an exercise in my text, I'm trying to show that $B = [0,1] \setminus A$ doesn't have any open interval $(x,y)$ with $0 \leq x < y \leq 1$. The prompt is to first show that for any $0 \leq x < y \leq 1$ there is a non-empty subinterval $(f,g) \subseteq (x,y)$ such that $(f,g) \subseteq A$. 
I'm having a hard time figuring out how to go about proving the prompt first of all. My first thought was to use the fact that between any two rationals/irrationals there is an irrational/rational and place $3$s on the end of decimal expansions to show that there must be some matching subset for that criteria, but I'm seriously doubting whether this approach is correct. 

Divide the interval $(0,1)$ into 10 subintervals of equal length: these sub-intervals will have end points of the for $k/10$ with $k=0,1,\ldots,10$.
All numbers of the fourth interval $(3/10,4/10)$ will have digit 3 as the first digit in the decimal expansion.
Now each of the 10 sub-intervals are further subdivided into 10 parts of equal length. In each of them consider the fourth subinterval  For example when $(7/10,8/10)$ is subdivided again fourth division would be $(73/10,74/10)$. The real numbers in each of these sub intervals will have 3 as the second digit in decimal expansion.
If we continue further in $n$th stage we will get intervals of length $1/10^k$.
Given an interval of length $\epsilon$ find $k$ such that $1/10^k$ is less than $\epsilon$. Then in $k$-th stage one of this sub interval will intersect the given interval.
