I've checked by hand that for any $n \le 7$, there is a cyclic subgroup $C$ of $S_n$ such that the intermediate subgroups lattice $\mathcal{L}(C \subset S_n)$ is distributive.
Question: Is it the same for $S_8$? and next for any $S_n$?
I've tried to check with GAP but after two days, the computation is not yet finished.
Is there a way to access to the atlas of subgroups lattice of Thomas Connor and Dimitri Leemans with GAP, or at least to their algorithm translated in GAP (it is now written for MAGMA)?
From the comments, we can assume that $N_{S_8}(C)/C$ does not contain a Klein $4$-group.
In fact that leaves only two possible $C$:
I can rule out the first one.
Let $H = \langle (1,2,3,4,5,6,7),(2,7)(3,6)(4,5) \rangle$ be dihedral of order $14$.
Let $K = \langle (1,2,3,4,5,6,7),(1,7)(2,5)(3,8)(4,6) \rangle$ be a Frobenius group of order $56$.
Then you can check that $\langle H,K \rangle = S_8$, so $A_8 \cap \langle H,K \rangle = A_8$.
But $A_8 \cap H = C$ and $A_8 \cap K = K$, so $\langle A_8 \cap H, A_8 \cap K \rangle = K$, and the lattice is not distributive.
I can now also rule out the $C_{10}$ case with $C = (1,2,3,4,5)(6,7) \rangle$. Consider the following three subgroups, all containing $C$:
(i) $H_1 = S_7$, the stabiliser of $8$ in $S_8$.
(ii) $H_2 = S_5 \times S_3$ with orbits $\{1,2,3,4,5\}$ and $\{6,7,8\}$.
(iii) $H_3 = S_6 \times S_2$ with orbits $\{1,2,3,4,5,8\}$ and $\{6,7 \}$.
Then any two of these generate $S_8$, but the intersection of any two of them is the same subgroup $S_5 \times C_2$, so we have one of the standard non-distributive sublattices.