Since $A$ is analytic, there exists a continuous surjection $f: \omega^\omega \to A$ (or we can write $A = p[\mathcal{T}]$ for a tree $\mathcal{T}$). We use the Cantor-Bendixson analysis combined with the tree representation.

Step 1: Tree representation. Write $A$ as the projection of a closed set $F \subseteq X \times \omega^\omega$: $A = \pi_X[F]$. This $F$ can be represented by a tree $\mathcal{T}$ on $X^{<\omega} \times \omega^{<\omega}$: $F = [\mathcal{T}]$ (the set of infinite branches).

Step 2: The Suslin scheme. Define a Suslin scheme $(F_s)_{s \in \omega^{<\omega}}$ by setting $F_s = \{x \in X : \exists y \in \omega^\omega,\, y \text{ extends } s,\, (x,y) \in F\}$. Then $A = \bigcup_{n} \bigcap_{k} F_{(n_0, \ldots, n_k)}$ where the union-intersection is over branches.

Step 3: Splitting. An uncountable analytic set has a splitting tree: a perfect subtree $S \subseteq \omega^{<\omega}$ such that for every node $s \in S$, the set $F_s \cap A$ is uncountable, and $s$ has at least two children $s^\frown 0, s^\frown 1 \in S$ with $F_{s^\frown 0}$ and $F_{s^\frown 1}$ both uncountable.

The construction proceeds by induction: at each level, use the fact that if $A_s = F_s \cap A$ is uncountable, it cannot be the union of countably many singletons. Since each $A_s = \bigcup_n A_{s^\frown n}$, at least one (in fact two, by a pigeonhole argument) must be uncountable.

Step 4: Extracting a perfect set. The splitting tree $S$ determines a continuous injection $\varphi: 2^\omega \to A$: for each $x \in 2^\omega$, follow the corresponding branch of $S$ to obtain a unique point $\varphi(x) \in A$. The image $\varphi[2^\omega]$ is a perfect set contained in $A$.

More precisely: refine the neighborhoods so that $\mathrm{diam}(F_s) \to 0$ along each branch (possible since $X$ is metrizable). Then each branch of $S$ determines a unique point in $A$, and the resulting set is homeomorphic to $2^\omega$ (the Cantor set), hence perfect. $\blacksquare$