Let $$X = \{(x,y) \in \mathbb{R}^2 : y = \pm 1\}$$ and let $M$ be the quotient of $X$ by the relation generated by $(x, -1) \sim (x,1)$ for all $x \neq 0$, i.e. $M$ is the line with two origins. I would like to show that $M$ is locally Euclidean and second-countable, but not Hausdorff.
My attempt: We may identify the points that do not intersect the double origin with $\mathbb{R}\setminus\{0\}$, which has a basis of open balls with rational centers. To account for the two origins, in addition to the open balls, we may consider sets of the form: $$A_\epsilon = \{(-\epsilon, 0) \cup \{(0, 1)\} \cup (0, \epsilon)\}, \quad B_\epsilon = \{(-\epsilon, 0) \cup \{(0, -1)\} \cup (0, \epsilon)\}$$ in other words, we split a ball of radius epsilon and attach either the point $(0,1)$ or $(0, -1)$. Second countability follows easily by taking the union of all such sets for various $\epsilon$ along with the balls with rational centers. Further, notice that for any $\epsilon$, $A_\epsilon \cap B_\epsilon \neq \emptyset$, and so the space fails to be Hausdorff.
All that remains to be shown is the locally Euclidean property, and so for each open $U \subset M$ we look for a homeomorphism $\varphi: U \rightarrow \tilde{U} \subset \mathbb{R}$. For open sets that do not intersect either origin we may take $\varphi$ to be the identity function.
I am struggling to define $\varphi$ on open sets that cross either origin. An old StackExchange question suggests identifying $A_\epsilon$ and $B_\epsilon$ with $(-\epsilon, \epsilon)$, but wouldn't this void injectivity?
Is the rest of my proof correct? I fear I am lacking some rigor in my arguments.