Problem. Show that for all nonzero $v\in\mathbb{R}^n$ and all $\epsilon>0$, there exists $r>0$ such that $\operatorname{dist}(rv,\mathbb{Z}^n- \{0\})<\epsilon.$
In other words, we must show that for all $\epsilon>0$ there exists $w\in\mathbb{Z}^n-\{0\}\subseteq\mathbb{R}^n$ such that $|v-w|<\epsilon$.
Bonus question. Can we take $w$ to be arbitrarily large?
Attempts. The case $n=1$ is trivial: we can take $r=1/v$ so that $rv=1$. But that approach obviously doesn't generalize to $n>1$. Any idea?
More generally, if $v$ is rational, say $v_i=a_i/b_i$, then we can take $r=b_1\cdots b_n$ so that $rv\in\mathbb{Z}^n-\{0\}$.
What you refer to as the 'dist()' function is more commonly referred to as the vector norm, a function which takes a vector and produces the 'length' of a vector. When you scale a vector by a constant, the norm of the vector scales by that constant (at least for the usual norm in this case). It follows that the argument given for n=1 can be used nearly verbatim, all you need to do is translate from scaling the vector to scaling the norm.