I was trying to prove the following: any initial ordinal is a member of the class $\mathscr{I} = \omega_0\cup\{\omega_\alpha:\alpha\in\mathscr{O}\}$, where $w_0$ is just $\mathbb{N}$ when viewed as an ordinal, $\mathscr{O}$ is the class of all ordinals. The ordinals $\omega_1$, $\omega_2$ and so on are constructed using transfinite recursion, where $\omega_{\alpha + 1}$ is the Hartog's number of $\omega_\alpha$ (this is just to give some motivation regarding the problem I am about to ask).
While writing the proof, I reached a step where I had to prove the following: if $\alpha,\beta$ are ordinals such that $\alpha\in\beta$, then $\beta$ cannot be order isomorphic to any subset of $\alpha$. I tried proving this using transfinite induction but couldn't handle all cases. It seems like this is true: for example, $\omega_0 + 1$ cannot be order isomorphic to any subset of $\omega_0$.
I know the following fact: two ordinals are order isomorphic if and only if they are equal, and one ordinal cannot be order isomorphic to an initial segment of itself. However, I am not able to put these together in proving what I asked.
Is this true, and can you guys help me with it?
Edit: If this is not true, maybe this is true: if $\alpha,\beta$ are ordinals such that $\alpha\in\beta$ and $\beta$ is an initial ordinal, then $\beta$ cannot be order isomorphic to any subset of $\alpha$. But I haven't been able to prove this either.