I'm trying to construct a proof that there is no homeomorphism from a subspace of $\omega_1$ to $\Bbb{Q}$ with the usual topology, but first I need to clear up some concepts.
This is the definition I am given:
$\alpha \in \omega_1$ is called a successor if $\operatorname{pred}(\alpha) \setminus \{\alpha\}$ has a maximal element. If $\alpha$ is not a successor then it is called a limit.
First I wan't to show that $\{\alpha\}$ is open if it is a successor and not open if it is a limit. Let $\{\alpha\}$ be a successor element, then $\operatorname{pred}(\alpha) \setminus \{\alpha\}$ has a maximal element, lets denote it by $\beta$, Now let $\alpha + 1$ denote the next least element after $\alpha$ then $(\beta, \alpha + 1)$ is an open set containing only $\{\alpha\}$ so it is open.
Let $\{\alpha\}$ now be a limit ordinal and let $\beta$ be an element less than $\alpha$, then $(\beta, \alpha + 1)$ is an open set containing $\alpha$, but since $\operatorname{pred}(\alpha)$ has no maximal element, we can always find a $y$ s.t $\beta < y < \alpha$ and $(y, \alpha + 1)$ is a strict subset of $(\beta, \alpha + 1)$ containing $\alpha$ and so $\{\alpha\}$ is not open.
Now for the actual question, let $Y$ be a countable subset of $\omega_1$, (if its not countable then we can't have a bijection with $\Bbb{Q}$), for $ y \in Y$ define $I_y = \{x \in Y: x \leq y \}$. Now consider $\{\min(Y \setminus I_y)\}$ then this is a successor ordinal and hence, open, but the set of $\Bbb{Q}$ with the usual topology has no open singletons so we have a contradiction.
Your basic idea is fine, but what you’ve written doesn’t quite work, because although $\min(Y\setminus I_y)$ is a successor in the relative order on $Y$, it need not be a successor in $\omega_1$. Thus, you need a bit more than your preliminary lemma: you need to show that if $Y\subseteq\omega_1$, and $\alpha\in Y$, then $\{\alpha\}$ is open in $Y$ if and only if $\alpha$ is a successor in the relative order on $Y$, i.e., if and only if $\operatorname{pred}_Y(\alpha)\setminus\{\alpha\}$ has a maximal element, where
$$\operatorname{pred}_Y(\alpha)=\{\beta\in Y:\beta\le\alpha\}=\operatorname{pred}(\alpha)\cap Y\;.$$
The proof of this is very similar to what you did for the case $Y=\omega_1$, so I’ll leave it to you for now to see if you can fill in the details for yourself.