Embedding the Schreier spaces into $C(\alpha)$ for some countable ordinal $\alpha$ Let $1\leq\xi<\omega_1$ be any countable ordinal, and denote by $\mathcal{S}_\xi$ the Schreier family of order $\xi$.  Then the Schreier space of order $\xi$ is the completion $S_\xi$ of $c_{00}$ under the norm
\begin{equation}\|(a_n)_{n=1}^\infty\|_{S_\xi}=\sup_{F\in\mathcal{S}_\xi}\|(a_n)_{n\in F}\|_{\ell_1}.\end{equation}
When $\xi=1$, the space $S_1$ is known to be $c_0$-saturated (every infinite-dimensional closed subspace contains a further subspace isomorphic to $c_0$).  The argument goes as follows:  $S_1$ embeds (via a continuous linear map) into $C(\omega^\omega)$.  But $\omega^\omega$ is (presumably?) a scattered, compact Hausdorff space, and so by the "main theorem" in the paper "Spaces of Continuous Functions (III)" by Pelczynski/Semadeni, $C(\omega^\omega)$ is $c_0$-saturated.  It follows that $S_1$ is $c_0$-saturated.
Question 1.  Does this argument work for $S_\xi$ when $1\leq\xi<\omega_1$ is an arbitrary countable ordinal?  In other words, if $1\leq\xi<\omega_1$, can we always find a countable ordinal $\alpha$ which is compact Hausdorff and scattered, such that $S_\xi$ embeds into $C(\alpha)$?
Question 2.  How do we know that $S_1$ linearly embeds into $C(\omega^\omega)$?  More generally, if the answer to question 1 is "yes," then how do we know that $S_\xi$ linearly embeds into $C(\alpha)$?
Thank you!
 A: Question 1. The quickest way to embed $S_1$ into a $C(K)$-space with $K$ countable (hence scattered) is to go via this construction. 
Let $L=\{-1,0,1\}^\mathbb{N}$. This is a compact Hausdorff space homeomorphic to the Cantor set. Define $$K = L / \{(\xi_n)_{n=1}^\infty\colon \text{support of }(\xi_n)_{n=1}^\infty \text{ is admissible}\}.$$
Note that $K$ is countable. Consider now $T\colon S_1\to C(K)$ given by
$$(Tx)([(\xi_n)_{n=1}^\infty]) = \sum_{n=1}^\infty \xi_n x_n\qquad ((\xi_n)_{n=1}^\infty\in L, x=(x_n)_{n=1}^\infty\in S_1).$$
We have
$$\|Tx\| = \sup_{[(\xi_n)_{n=1}^\infty]\in K}|\sum_{n=1}^\infty \xi_n x_n|=\sup_{A\,{\rm admissible}}\sum_{n\in A}|x_n|=\|x\|_{S_1}\qquad (x\in S_1).$$
With some effort you can compute the Cantor–Bendixson rank of $K$ and use the Bessaga–Pełczyński classification theorem to see that $C(K)\cong C(\omega^\omega)$.
Question 2. Yes, the higher order Schreir spaces embed into $C(K)$ with $K$ countable too. You can mimic the above construction. However, even more is true–see this paper by Gasparis.
