In fact, you can prove directly that any countable compact space $X$ is metrizable:
Let $A$ be a family of continuous functions from $X$ to $\mathbb{R}$ and let $e : \left\{ \begin{array}{ccc} X & \to & \mathbb{R}^A \\ x & \mapsto & (f(x)) \end{array} \right.$. If the family $A$ distinguishes points and closed sets in $X$, then $e$ embedds $X$ into $\mathbb{R}^A$; since $X$ is completely regular, such a family $A$ exists. Taking the linear subspace spanned by $e(X)$, you can suppose $A$ countable.
Therefore, $X$ is a subspace of $\mathbb{R}^{\omega}$; by compactness, $X$ is in fact a subspace of a product of intervals $\prod\limits_{\omega} I_k$. Since each $I_k$ is homeomorphic to $[0,1]$, $X$ is a subspace of $[0,1]^{\omega}$.
But $[0,1]^{\omega}$ is metrizable, $d : (x,y) \mapsto \sum\limits_{n \in \omega} \frac{1}{2^n}|x_n-y_n|$ defines a distance over it.
Moreover, since any metrizable space can be embedded into a normed space, you can use the proof given in [Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920] to prove the classification you mention:
First notice that the Cantor-Bendixson rank of a countable compact space is a successor countable ordinal $\alpha+1$ and that $X_{\alpha}$ is finite. Then for any countable ordinal $\alpha$ and positive integer $n \geq 1$, let $P(\alpha,n)$ be the assertion: "Any countable compact space of Cantor-Bendixson rank $\alpha+1$ such that $\text{card}(X_{\alpha})=n$ is homeomorphic to $\omega^{\alpha} \cdot n+1$."
Clearly, $P(1,1)$ is true (here $X$ is just a converging sequence).
Step 1: If $P(\alpha,1)$ is true then $P(\alpha,n)$ is true.
Let $X_{\alpha}=\{p_1, \dots,p_n\}$. Then, viewing $X$ as a subspace of a normed space $Y$, there exist $n-1$ parallel hyperplanes $P_1$, ..., $P_{n-1}$ such that $Y \backslash \bigcup\limits_{i=1}^{n-1} P_i$ has $n$ connected components $D_1$, ..., $D_n$ with $p_k \in D_k$.
Because $P(\alpha,1)$ is true, each $X_k:= X \cap D_k$ is homeomorphic to $\omega^{\alpha}+1$. Therefore, $X$ is homeomorphic to $(\omega^{\alpha}+1) \cdot n= \omega^{\alpha} \cdot n+1$.
Step 2: If $P(\alpha,n)$ is true for any $\alpha<\alpha_0$ and $n \geq 1$, then $P(\alpha_0,1)$ is true.
Let $X_{\alpha_0}=\{p\}$ and let $(p_k)$ be a sequence in $X'$ converging to $p$ (without loss of generality, we suppose $\alpha_0 \geq 2$ so that $p \in X''$). Then, viewing $X$ as a subspace of a normed space $Y$, there exist a sequence of postive real numbers $(r_k)$ converging to zero such that the family of spheres $S(p,r_k)$ does not meet $X$ and $Y \backslash \bigcup\limits_{k \geq 1} S(p,r_k)$ has infinitely many connected components $D_1$, $D_2$, ... with $p_k \in D_k$.
By assumption, each $X_k:= X \cap D_k$ is homeomorphic to some $\omega^{\alpha_k} \cdot n_k+1$ with $\alpha_k <\alpha_0$ and $n_k \geq 1$. Therefore, $X$ is homeomorphic to $\tau=[(\omega^{\alpha_1} \cdot n_1+1)+(\omega^{\alpha_2} \cdot n_2+1)+ \dots ]+1$
But $\tau \leq \omega^{\alpha_0} +1$. If $\tau < \omega^{\alpha_0}+1$ then $\tau< \omega^{\alpha_0}$ because $\tau$ is compact, hence $X_{\alpha_0}= \emptyset$: a contradiction. Therefore, $\tau = \omega^{\alpha_0}+1$ and $P(\alpha_0,1)$ is true.
Step 3: We conclude by transfinite induction.