Suppose $b$ is irrational. I will first prove the result when $a=1$. For each positive integer $m$ there exists a unique integer $n_m$ in the interval $[-mb,-mb+1)$. Let $x_m=n_m+mb$ so $x_m \in [0,1)$ for each $m$. Choose a positive integer $k$ such that $\frac 1 k <\epsilon$. Consider the number $x_1,x_2,...,x_{k+1}$ in $[0,1)$. If any two of these differ by $\frac 1 k$ or more then the difference between the largest and the smallest of them would exceed $1$ which is a contradiction. In other words there exist $i,j$ such that $0<|x_i-x_j| <\frac 1 k <\epsilon$. Thus $0<|(n_i+ib)-(n_j+jb)| <\epsilon$. [The fact that $b$ is irrational is required to say that $(n_i+ib)-(n_j+jb) \
eq 0]$. This proves the result when $a=1$. A simple modification gives the result for any $a$.