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.