Jackson's Subsets

Let $\mathbb S=\left(-\pi/2,\pi/2\right)$ . The function $\tan : \mathbb S\mapsto \mathbb R$ ranges over all real numbers, so for all real numbers $x_i$ there exists a unique $\theta_i\in\mathbb S$ such that $x_i=\tan\theta_i$ . So there exists a bijection between sequences of real numbers with their respective $\theta_i\in\mathbb S$ value. We can hence work only with the $\theta_i$ values.

Suppose $x_a=\tan \theta_a$ and $x_b=\tan\theta_b$ , and WLOG let $\theta_a\ge\theta_b$ . We have $\tan\dfrac\pi 6=\dfrac{1}{\sqrt 3}\ge \left|\dfrac{x_a-x_b}{1+x_ax_b}\right|=\left|\dfrac{\tan\theta_a-\tan\theta_b}{1+\tan\theta_a\tan\theta_b}\right|=\left|\tan\left(\theta_a-\theta_b\right)\right|$ and since $\theta_a\ge\theta_b$ we have $\tan\dfrac\pi 6\ge\tan\left(\theta_a-\theta_b\right)\implies \theta_a-\theta_b\le \dfrac\pi 6$ because $\tan:\mathbb S\mapsto \mathbb R$ is increasing.

Now split $\mathbb S=\left(-\dfrac\pi 2,\dfrac\pi 2\right)$ into the following $6$ intervals $\left(-\dfrac {\pi} 2,-\dfrac{\pi} 3\right],\left(-\dfrac {\pi} 3,-\dfrac\pi 6\right],\left(-\dfrac \pi 6,0\right],\left(0,\dfrac\pi 6\right],\left(\dfrac\pi 6,\dfrac \pi 3\right],\left(\dfrac\pi 3,\dfrac\pi 2\right).$ If $n\le 6$ we can choose all the $\theta_i$ from different intervals, equally spaced, so that we'll have $\theta_i-\theta_j>\pi/6$ for all $i,j$ . Therefore all sequences with $n\le 6$ might not satisfy the problem conditions.

If $n=7$ then by the Pigeonhole Principle at least two $\theta_i$ will be from the same interval, so we'll have $\theta_a-\theta_b\le\pi/6$ for those two values. Hence $n=\fbox{7}$ is the smallest size of such sequence.