 Solution

For subgroups $H,K\le G$ and an $H$-representation $W$, Mackey restriction formula isFrobenius reciprocity statesApplying these with $K=H$ givesThe identity double coset contributes one exactly when $W$ is irreducible, and every summand is a nonnegative integer. This proves Mackey irreducibility criterion: the induced representation is irreducible exactly when $W$ is irreducible and every nonidentity double-coset summand vanishes.

Now take $G=S_n$, $H=S_{n-1}$, and $x=(n-1n)\notin H$. Thenthe subgroup fixing both $n-1$ and $n$. Conjugation by $x$ fixes every element of this subgroup, so the two representations appearing in the corresponding Mackey summand are both ${\mathrm{Res}}_{S_{n-2}}^{S_{n-1}}W$. Their inner product is positive because this restriction is nonzero. Hence the Mackey irreducibility criterion fails whenever $W$ is irreducible. If $W$ is reducible, induction preserves its direct-sum decomposition, so the induced representation is again reducible. Therefore ${\mathrm{Ind}}_{S_{n-1}}^{S_n}W$ is never irreducible for $n\ge 2$.