= Solution
The <symmetric group> is transitive on two-element subsets, giving $\langle\mathbf1,\chi_Y\rangle=1$. Its diagonal action on $X\times Y$ has two orbits, according to whether the point lies in the pair. Thus $\langle\chi_X,\chi_Y\rangle=2$, and, since $\chi_X=\mathbf1+\chi_S$, the standard zero-sum <representation> $S$ occurs in $\mathbb C[Y]$ once. Its irreducibility follows from the double transitivity of the natural symmetric-group action and the corrected statement of part (ii).
For $n\ge4$, ordered pairs of two-element subsets have three possible intersection sizes, zero, one and two, each forming a single orbit. Part (i) therefore gives $\|\chi_Y\|^2=3$. After removing the one copy each of $\mathbf1$ and $S$, the remaining <character> has squared norm $3-1-1=1$, so is itself irreducible. Hence
$$
\boxed{\mathbb C[Y]\cong\mathbf1\oplus S\oplus R,\qquad\dim R=\binom n2-1-(n-1)=\frac{n(n-3)}2\quad(n\ge4).}
$$
For $n=3$, taking the complement of a pair is an equivariant bijection $Y\to X$. Thus \b[$\mathbb C[Y]\cong\mathbf1\oplus S$], with no third constituent. Correspondingly the disjoint-pair orbit is absent and the <character> norm is two.
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