= Solution
For subgroups $H,K\leq G$ and an $H$-representation $W$, <Mackey restriction formula> is
$$
\operatorname{Res}_K^G\operatorname{Ind}_H^G W
\cong
\bigoplus_{x\in K\backslash G/H}
\operatorname{Ind}_{K\cap xHx^{-1}}^K
\operatorname{Res}_{K\cap xHx^{-1}}^{xHx^{-1}}({}^xW).
$$
<Frobenius reciprocity> states
$$
\langle\operatorname{Ind}_H^G\chi,\psi\rangle_G
=\langle\chi,\operatorname{Res}_H^G\psi\rangle_H.
$$
Applying these with $K=H$ gives
$$
\left\langle\operatorname{Ind}_H^G\chi,
\operatorname{Ind}_H^G\chi\right\rangle_G
=
\sum_{x\in H\backslash G/H}
\left\langle
\operatorname{Res}_{H\cap xHx^{-1}}^H\chi,
\operatorname{Res}_{H\cap xHx^{-1}}^{xHx^{-1}}({}^x\chi)
\right\rangle.
$$
The 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 $x=(n-1\ n)\notin H$. Then
$$
H\cap xHx^{-1}=S_{n-2},
$$
the 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 $\operatorname{Res}_{S_{n-2}}^{S_{n-1}}W$. 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 $\operatorname{Ind}_{S_{n-1}}^{S_n}W$ is never irreducible for $n\geq2$.
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