Let $(x_n)_n$ be a sequence of distinct points in $\mathbb{R}^2.$ Is there a straight line $l\subset \mathbb{R}^2$ and a subsequence $(x_{k_n})_n$ of $(x_n)_n$ such that $d(x_{k_n},l)\overset{n\to\infty}{\to} 0\ ?$
If $(x_n)_n$ is bounded or has limit points then the answer is trivially yes by Bolzano-Weierstrass. So suppose $(x_n)_n$ is not bounded and has no limit points.
If we try fixing a point and require the line go through that point, then argue that there are a sequence of angles $(\theta_k)_k$ relative to the positive horizontal about that point such that there are infinitely many $x_n$ in the region bounded between $\theta_k$ and $\theta_{k+1}$ about that point, with $\theta_{k-1}<\theta_{k+1}< \theta_k$ and $\lim_{k\to\infty} ( \theta_k - \theta_{k+1}) = 0,$ then this gives us a straight line: $\lim_{k\to\infty} \theta_k.$ However, I think this argument fails to give us a line where $d(x_{k_n},l)\overset{n\to\infty}{\to} 0$ via Bolzano-Weierstrass, because for large $k$, the points $x_n$ could be very far away from the fixed point, and therefore far away from the line in the limit, and so it isn't clear this argument works.
So, any hints or ideas?
Consider the sequence $x_n=(n,n^2)$ as part of the standard parabola. It's clear that vertical lines won't work. But any non-vertical lines will eventually be caught up by the parabola (when going to $+\infty$). But since the sequence consists out of distinct points any subsequence will eventually converge much faster to infinity away from the line; in particular we cannot have $d(x_{k_n},l)\rightarrow 0$.