TITLE: If a subset of the plane has open intersection with every line, is it open?
QUESTION [9 upvotes]: Suppose that $U \subset \mathbb{R}^2$ is such that $U \cap L$ is open in $L$ for any line $L \subset \mathbb{R}^2$ where $L$ inherits the subspace topology from $\mathbb{R}^2$ (ie. $L \cong \mathbb{R}$). Does it follow that $U$ is open? I keep thinking I have a counterexample and then changing my mind...
Not sure if the topological vector space tag is appropriate. I thought if the question had a positive answer then it might have something to do with the fact that every finite dimensional vector space has a unique Hausdorff topology compatible with the operations.

REPLY [6 votes]: Let $$p_n = \left\langle \cos\frac{\pi}{2^{n+1}},\sin\frac{\pi}{2^{n+1}} \right\rangle,$$ let $P=\{p_n:n\in\omega\}$, and let $S=\mathbb{R}^2\setminus P$. The intersection of any line with $P$ is either a singleton or a doubleton, so every line intersects $S$ in an open set, but $\langle 1,0 \rangle \in S \cap \operatorname{cl}P$, so $S$ isn't open.
Any convergent sequence with no three points collinear will do.