How many non-homeomorphic subspaces of $\mathbb Q$ are there?
$\aleph_1$ ? $\mathfrak c$ ? Is it known?
EDIT.
Since the time I originally posted the question I found out about the result in https://arxiv.org/abs/1504.08130 (Proposition 19). I haven't studied it yet, but it solves my question, because each countable ordinal is emeddable into $\mathbb Q$.
Now I am leaving the question for 'share knowledge' reasons as there are possibly other ways of proving this (one of them suggested by Eric Wofsey).
As you observe, every countable ordinal embeds topologically (indeed, order-theoretically) into the rationals. Distinct ordinals can be homeomorphic (e.g. $\omega+1\cong\omega+2$), but distinct ordinals of the form $\omega^\alpha$ for some $\alpha$ (= "right-indecomposable") aren't, so this gives us $\aleph_1$ many homeomorphism types. But contra your edit, this leaves open the issue of whether there are as many as possible (= $\mathfrak{c}$-many) in case CH fails.
To get continuum-many, we can use the following trick. First fix a closed $A\subset\mathbb{Q}$ of ordertype $\omega^\omega$; note that $A$ has limit points of every finite order (limit, limit-of-limits, limit-of-limits-of-limits, etc.), and let $\alpha_i$ be an element of $A$ which is an $i$-fold limit but not an $(i+1)$-fold limit for each $i$. For $p\in A$, let $\epsilon_p$ be half the distance between $p$ and the next element of $A$ (which is positive since $A$ is well-ordered).
Given $X\subseteq\mathbb{N}$, let $$A_X=A\cup\bigcup_{i\in X}([\alpha_i,\alpha_i+\epsilon_{\alpha_i}]\cap\mathbb{Q}).$$ Basically, we attach a "$\mathbb{Q}$-like label" to the $\alpha_i$s which code elements of $X$, and don't attach these labels to the ones that don't. It's a good exercise to show that $A_X\cong A_Y\iff X=Y$, and so we do indeed get continuum-many pairwise-non-homeomorphic subspaces of $\mathbb{Q}$.