As Amritanshu Prasad points out, the $6$-dimensional irreducible complex representation of $S_{5}$ is indeed monomial (with respect to a suitable basis), and thinking about how to prove this directly led me to a general observation: let $G$ be a finite group, and $\chi$ be a non-linear complex irreducible character of $G$ of minimal degree. Let $H$ be a proper subgroup of $G$ of index less than $2\chi(1).$ Let $\lambda$ be a linear character of $H$ of order $m$ ( that is, $\lambda^{m}$ is the trivial character, but no smaller positive integer power of $\lambda$ is trivial). Then if $m$ does not divide $[G:G^{\prime}],$ the induced character $\theta = {\rm Ind}_{H}^{G}(\lambda)$ is irreducible. For otherwise, (by Frobenius reciprocity), there must be a linear constituent $\mu$ of $\theta$ such that $\langle {\rm Res}^{G}_{H}(\mu),\lambda \rangle \neq 0.$ But then the order of $\mu$ must be divisible by $m,$ whereas the order of $\mu$ divides $[G:G^{\prime}],$ a contradiction.

To apply this result to $G = S_{5},$ take $H$ to be the normalizer of a Sylow $5$-subgroup. Then $[G:G^{\prime}] = 2,$ and $H$ has a linear character $\lambda$ of order $4,$ so the above result can be used to see that ${\rm Ind}_{H}^{G}(\lambda)$ is irreducible (for note that $S_{5}$ has no non-linear irreducible character of degree less than $4$).

Incidentally, taking $G = A_{5},$ and $K$ to be a Sylow $5$-normalizer of $G$, we see that $[G:G^{\prime}] = 1,$ that $[G:K] = 6$ and that $G$ has no non-linear irreducible character of degree less than $3.$ However, $K$ has a linear character of order $2$ such that ${\rm Ind}_{K}^{G}(\lambda)$ is not irreducible ( it is a sum of two irreducible characters of degree $3$). Hence the bound $[G:H] < 2 \chi(1)$ in the result above can't be sharpened in general.