Option A

$S_5$ contains a cyclic subgroup of order $6$. Consider the subgroup generated by $\sigma = (12)(345)$. The order of $\sigma $ is the least common multiple (LCM) of the lengths of cycle (as they are disjoint), which is $6$. Thus the subgroup $\left\langle \sigma \right\rangle $ is a cyclic subgroup of order $6$. Therefore, this statement is true.

Option B

In $S_5$, we can see a copy of $D_4$, the dihedral group (symmetries of square. In some places they use the notation $D_8$). Since $D_4$ is non-abelian and has order $8$, we conclude that $S_5$ contains a non-abelian subgroup of order $8$. To see this, take any four symbols, say $1,2,3,4$ and label the vertices of the square by $1,2,3$ and $4$. Then \[ D_4 = \left\langle (1234), (13) \right\rangle . \] More precisely, $\left\langle (1234) \right\rangle = \{ (1), (1234), (13)(24), (1432) \} $ is the rotation subgroup and the reflections are $(13), (24), (12)(34)$ and $(14)(23)$. Therefore, the group $S_5$ has a subgroup of order $8$. This is true in general, that is, $D_n$ is a subgroup of $S_n$ (see here).

Option C

The order of $S_5$ is $5! = 120$ and hence from Lagrange's theorem order of any subgroup of $S_5$ must divides $120$. Since $7$ does not divide $120$, there can not be any subgroup of order $7$.

Option D

The order of the group $\mathbb{Z} /2\mathbb{Z} \times \mathbb{Z} /2\mathbb{Z} $ is $4$ and it is not-cyclic. Consider the subgroup of $S_5$ defined as \[ H = \{ (1), (12), (34), (12)(34) \}. \] This subgroup is isomorphic to $\mathbb{Z} /2\mathbb{Z} \times \mathbb{Z} /2\mathbb{Z} $.