The line with two origins is $(\mathbb{R} \times \{0,1\})/\sim$ where $(x,0)\sim(x,1)$ for $x\neq 0$. I can see that it is not Hausdorff, since we cannot separate the points $(0,0)$ and $(0,1)$.
However, I'm not quite clear on why it is a manifold, in particular why it is locally Euclidean. Please help me understand why this is so.
Best Answer
The key here is what is the topology on this space?
A possible base for this topology is $\mathcal B$ which contains all open balls not containing zero, and at zero, we either have $\{(-\varepsilon, 0) \cup \{\text{origin } A\} \cup (0, \varepsilon)\}$ and $\{(-\varepsilon, 0) \cup \{\text{origin } B\} \cup (0, \varepsilon)\}$.
Note: If you restrict the basis to rational endpoints, we get a countable basis, making the space second countable, which might also be in your definition of manifold.
Consider for instance $\{(-\varepsilon, 0) \cup \{\text{origin } A\} \cup (0, \varepsilon)\}$. There is an obvious bijection to $(-\varepsilon, \varepsilon)$ which is continuous and whose inverse is continuous.