Let $\Omega$ be the first non-numerable ordinal number ( $\aleph_1$ is the first cardinal number greater than $\aleph_0$ when treated as an ordinal number is denoted by $\Omega$ ) and let $[0,\Omega)$ be the subspace of the ordinal space $[0,\Omega]$. Then every continuous $\varphi : [0,\Omega)\rightarrow E^{1}$ must be constant in a tail of $[\beta,\Omega)$.
My attempt
It must be proved that the function $\varphi$ is constant from a certain value of the subspace $[0,\Omega)$, i.e., for all $n\in \mathbb{Z}^{+}$ there exists a $\alpha_n < \Omega$ for all $\xi> \alpha_n:|\varphi(\xi)-\varphi(\alpha_n)|< \frac{1}{n}$ (*).
Suppose this were not true, then there exists a $n_0$ for every $\alpha<\Omega$ there exists a $\xi>\alpha$ such that $|\varphi(\xi)-\varphi(\alpha)|< \frac{1}{n_0}$. Since each element of $[0, \Omega)$ has an immediate predecessor then, we can use induction to construct a countable succession, that is $\{\xi_i: i\in \mathbb{Z}^{+}\}$ such that. $$\xi_i<\xi_{i+1}$$ and $$|\varphi(\xi_i)-\varphi(\xi_{i+1})| \geq \frac{1}{n_0}$$ for each $i$, choosing $\xi_1=0$ and assuming that $\xi_1, \xi_2,...,\xi_k$ are defined, since $\xi_k$ has a successor, then $\xi_{k+1}$ is the first element in $[\xi_k, \Omega)$ that satisfies the hypothesis.
The succession $\{\xi_i\}$ thus constructed, would have a minimum upper bound $\gamma<\Omega$ and then $\varphi$ would not be continuous in $\gamma$, since any neighborhood $(\eta, \gamma]$ contains some $\xi_i$ and therefore any $\xi_k$, $k>i$ cannot have an image contained in the $$\left( \varphi(\gamma)-\frac{1}{3n_0}, \varphi(\gamma)+\frac{1}{3n_0}\right)$$ neighborhood of $\varphi(\gamma)$. Since this contradicts the definition of continuity, let's see.
Let $(\eta,\gamma]$ be an open containing $\gamma$. This open contains infinitely many $k_i$, for $k>i$ we have $$|\varphi(\xi_k)-\varphi(\gamma)|$$ but $$|\varphi(\xi_k)-\varphi(\xi_{k+1})+\varphi(\xi_{k+1})-\varphi(\gamma)|=|\varphi(\xi_k)- \varphi(\gamma)|\leq \frac{1}{3n_0}$$ luego $$|\varphi(\xi_k)-\varphi(\xi_{k+1})|-|\varphi(\xi_{k+1})-\varphi(\gamma)|\leq \frac{1}{3n_0}$$ $$\frac{1}{n_0}-|\varphi(\xi_{k+1})-\varphi(\gamma)|\leq \frac{1}{3n_0}$$ from where $$\frac{1}{n_0}\leq \frac{1}{3n_0}+\frac{1}{3n_0}$$ with which we have that $1\leq \frac{2}{3}$, this is a contradiction, therefore it is satisfied that for all $n\in \mathbb{Z}^{+}$ there exists a $\alpha_n < \Omega$ for all $\xi> \alpha_n: |\varphi(\xi)-\varphi(\alpha_n)|< \frac{1}{n}$.
Now, let $\beta$ be an upper bound of $\alpha_n$;then $\beta <\Omega$ and $\varphi$ is constant in $[\beta,\Omega)$: We must prove that if $\beta \in [\beta,\Omega)$, then the images are equal, otherwise we have that $$|\varphi(\zeta)-\varphi(\alpha_{n})|<\frac{1}{n}$$ and $$|\varphi(\beta)-\varphi(\alpha_n)|<\frac{1}{n}$$ for all $n$, thus. \begin{equation*} \begin{split} |\varphi(\zeta)-\varphi(\beta)|=&|\varphi(\zeta)-\varphi(\alpha_n)+\varphi(\alpha_n)-\varphi(\beta)|\\ \leq & |\varphi(\zeta)-\varphi(\alpha_n)|+|\varphi(\alpha_n)-\varphi(\beta)|\\ \leq & \frac{1}{n}+\frac{1}{n}\\ =&\frac{2}{n} \end{split} \end{equation*} for all n.
And therefore $\varphi(\zeta)=\varphi(\beta)$.
the proof is correct ? and I would also like to justify the step (*)
