= Solution
Work over the <complex numbers>. The chain $S_0\subset S_1\subset\cdots\subset S_n$ has <multiplicity-free restriction>: each <irreducible representation> restricts to a <direct sum> of pairwise inequivalent <irreducible representations>. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted <Olshanskii centralizer lemma> makes $C_{\mathbb C S_r}(\mathbb C S_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative <matrix algebra> factor. Thus no identification of the branching graph with <Young diagrams> is being assumed here.
Successively restricting an <irreducible representation> $V$ gives one-dimensional spaces $V_T$ indexed by paths of <irreducible representations> from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a <Gelfand–Tsetlin basis>. Define the <Gelfand–Tsetlin algebra> as the subalgebra of $\mathbb C S_n$ acting diagonally in all these bases. The <Artin–Wedderburn theorem> identifies
$$
\mathbb C S_n\cong\bigoplus_{V\in\operatorname{Irr}(S_n)}\operatorname{End}_{\mathbb C}(V).
$$
Under this identification the <Gelfand–Tsetlin algebra> is the direct sum of the full diagonal <matrix algebras> in the indicated bases, so it is commutative.
To see that it is actually available inside the <group algebra>, let $e_\alpha^{(r)}$ be the central <primitive idempotent> selecting the <irreducible representation> $\alpha$ of $S_r$. For a path $T=(\alpha_0,\ldots,\alpha_n)$ the product
$$
P_T=\prod_{r=0}^n e_{\alpha_r}^{(r)}
$$
is the projection onto $V_T$ in its final <irreducible representation> and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal <matrix units>, and therefore span the proposed algebra.
If an element of the <group algebra> commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the <Gelfand–Tsetlin algebra>. Thus
$$
\boxed{C_{\mathbb C S_n}(GZ_n)=GZ_n,}
$$
which proves that it is a <maximal commutative subalgebra>.
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