No conditions are needed: Divide the real line up into intervals of length $\epsilon$: $I_k=[k\epsilon,(k+1)\epsilon)$. Now you choose a point in $I_k$ with some probability $p_k=\int_{I_k}f$.

In order for $x_{n+1}$ to be at a distance $\epsilon$ from the other points chosen so far, it must be the first point to land in the interval $I_k$.

The probability of this happening is $\sum_{k\in\mathbb Z} p_k(1-p_k)^n$, so it is sufficient to show that for any probability distribution $(p_k)$ on $\mathbb Z$, the above sum converges to 0.

Notice that the $k$th term in the sum is monotonically decreasing to 0 and even for $n=0$ is integrable. (A very simple case of) the monotone convergence theorem implies that the sum converges to 0 as $n\to\infty$. Since this is an upper bound for what you want, you're done.