Let $0<r<1$ a real number which is not a fraction of the form $p/2^n$ for any integers $p,n$. Now, for every integer $n\ge 1$ we can find the closest fraction of the form $p/2^n$ to $r$, which will be on the left or on the right of $r$. For example, for $r=2/3$, those fractions will be: $1/2$, $3/4$, $5/8$, $11/16$ etc.
We will now look at the sequence $a_n=(3/2)^n\pmod 1$, which itself contains fractions of the form $p/2^n$ from the interval $(0,1)$, and could, for any given $r$, potentially "hit" one of those "closest" fractions to $r$.
Now, my question is this: can it be proven that, for every $r$, $0<r<1$, not of the form $p/2^n$, the sequence $a_n$ above contains at least one of those "closest" fractions to $r$? It seems intuitive to me that this should be true.
The fraction $a_1=3/2\pmod 1 = 1/2$ is the "closest" to all the numbers in $(1/4, 3/4)$. The fraction $a_2=9/4\pmod 1 = 1/4$ is the "closest" to all numbers in $(1/8, 3/8)$ etc. Generally, the fraction $a_n$ will be the closest to all the numbers in $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$. So the numbers $r$ for which one of $a_n$'s is the closest are precisely those contained in the union of those intervals $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$.
Now, the length of the interval $(a_n-1/2^{n+1}, a_n+1/2^{n+1})$ is $1/2^n$ so the total size (measure) of the union cannot exceed the sum of all of the lengths, which is $1/2+1/4+1/8+\ldots=1$. However, obviously, if any two of those intervals have a "nontrivial" intersection (however small, of size $\alpha>0$) the total measure of the union won't exceed $1-\alpha<1$.
Finally, it is enough to notice that the first two intervals $(1/4, 3/4)$ and $(1/8,3/8)$ already have a nontrivial intersection. Thus, the total measure of the union is strictly smaller than $1$, and so there must be a number $r$ which does not belong to this union. This $r$ will consequently never have any of the $a_n$'s as the "closest" to it.
In short, the statement you are trying to prove is not true.