We know that all connected (not a singleton) subsets of $\mathbb{R}$ (with the usual topology) has no empty interior. This fact does not remains true for a general connected topological space with the Lebesgue covering dimension equal 1 as showed in [Topological spaces with Lebesgue covering dimension 1](https://mathoverflow.net/questions/254854/topological-spaces-with-lebesgue-covering-dimension-1/254855#254855). However, the topological space presented in that post does not have a basis of connected sets. My question is: If $X$ is a connected topological space with Lebesgue covering dimension 1 having a basis of connected subsets then all connected subset (not a singleton) has no empty interior?



One-dimensional topological spaces
Unless I'm missing something, the answer is no: consider the space $\mathbb{R}'=\mathbb{R}\cup\{\alpha\}$, with topology generated by $$\{(a, b): a<b, 0\not\in (a, b)\}\cup\{(a, b)\cup\{\alpha\}: a<0<b\};$$ so we've duplicated $0$ to break $T\_0$-ness. Then $\mathbb{R}'$ still has Lebesgue dimension $1$ and a basis of connected sets, but $\{0, \alpha\}$ is a connected non-singleton subset of $\mathbb{R}'$ with empty interior.


Maybe you want to restrict to $T\_0$-spaces (or better)?


