Long ray: Proof of being locally euclidean Consider the topological space $X=\omega_1 \times [0,1)\setminus (0,0)$ equipped with the order topology that arises from the lexicographical order. I want to show that this space is locally euclidean. Clearly, for points $(\alpha,x)$ with $x>0$ I can construct a suitable homeomorphism. That's also no problem for points $(\beta,0)=(\alpha+1,0)$ where $\beta$ is a successor ordinal. But in the general case $(\lambda,0)$ where $\lambda$ is a limit ordinal, I have no idea. Nevertheless, for special limit ordinals like $\omega_0$ I worked out a proof but I don't know how to generalize it to all limit ordinals less than $\omega_1$.
Here is my homeomorphism for $p=(\omega_0,0)$:
We take $U=\{x\in X \mid x<(\omega_0+1,0)\}$ as open neighborhood of $p$.
$$f:U\to(1,3)$$
$$(n,x)\mapsto x\cdot 2^{-n-1}+\sum_{i=0}^n 2^{-i} \quad\text{for}\quad n\in\mathbb{N}$$
$$(\omega_0,x)\mapsto 2+x \quad\text{else}$$
I think, that map should do it.

The key remark is that $\omega_1$ is the first uncountable ordinal. Therefore, any $\omega<\omega_1$ is countable. You can adapt your proof as follows. Chose $\omega$ and for any $\alpha<\omega$ chose a positive number $\epsilon_\alpha$ such that the $\sum \epsilon_\alpha=1$. This is possible because $\omega$ is countable. Now, build your local homeo, as you did for $\omega_0$, by concatenating intervals of lengths $\epsilon_\alpha$:
$$(\alpha,x)\mapsto x\epsilon_\alpha+\sum_{\beta<\alpha}\epsilon_\beta$$
