For, $[a,b]$ in $\mathbb{R}$, $a,b$ are not interior points. However if I observe $([a,b] ,\{U\cap [a,b] : U\in \mathcal{T_\mathbb{R}}\})$ then if I understand correctly $a,b$ are interior points of $[a,b]$ as a space with the induced topology.
My confusion arises from the next problem - if $G$ acts PD upon Hausdorff, path connected and locally path connected space then $p:X\overset{p(x)=O(x)}\longrightarrow X / G $ is a regular covering space. Another theorem gives us that if $X$ is an n-dimensional manifold then $X/G$ is a n-dimensional manifold.
Combining those together, taking $G$ acts PD upon $\mathbb{R}$, I wished to claim that $\mathbb{R}/G$ can't be homeomorphic to $[a,b]$ because we got $\mathbb{R}$ as $\mathbb{R}/G$'s cover, and a covering map is a local homeomorphism then an interior point in $\mathbb{R}$ is mapped into a interior point in $[a,b]$, so $p$ can't map any point to $a,b$. Therefore $\mathbb{R}/G$ can't be homeomorphic $[a,b]$ but all this fail in case we are looking at $[a,b]$ with the induced topology where $a,b$ are interior points.
Am I suppose to observe $[a,b]$ with another topology? or maybe I have a deeper misunderstanding of the concepts I was trying to use? any explanation would be appreciated, also other ways to explain why $\mathbb{R} / G$ can't be homeomorphic to $[a,b]$