The statement that you want to prove follows from Dirichlet's Theorem, which says that for any real number $\alpha$ and for any positive integer $Q$, there is a fraction $a/q$ with $|q|\le Q$ such that $$|\alpha-a/q|\le 1/q(Q+1).$$ Here $\alpha$ is the slope of your line and $Q$ is playing the role of $R$, with $1/(Q+1)$ playing the role of $\epsilon$. Of course this depends, up to a constant, on what norms you are using. Also technically, to make the connection clearer, you should choose $\alpha$ to be the slope of the line or its reciprocal whichever is smaller.

For a paticular choice of $\mathbf{v}$ the fastest way to actually find the best integer vectors, as Igor pointed out, is by using the continued fraction algorithm. If you understand the connection to Dirichlet's Theorem then it should be clear how to do this.

The statement that you want to prove follows from Dirichlet's Theorem, which says that for any real number $\alpha$ and for any positive integer $Q$, there is a fraction $a/q$ with $|q|\le Q$ such that $$|\alpha-a/q|\le 1/q(Q+1).$$ Here $\alpha$ is the slope of your line and $Q$ is playing the role of $R$, with $1/(Q+1)$ playing the role of $\epsilon$. Of course in your problem the actual numerical value of $R$ as a function of $\epsilon$ depends, up to a constant, on what norms you are using. Also technically, to make the connection clearer, you should choose $\alpha$ to be the slope of the line or its reciprocal, whichever is smaller.

For a paticular choice of $\mathbf{v}$ the fastest way to actually find the best integer vectors, as Igor pointed out, is by using the continued fraction algorithm. If you understand the connection to Dirichlet's Theorem then it should be clear how to do this.