Let $G=D_{12}=\langle a,b \mid a^6=b^2=e, bab^{-1}=a^{-1} \rangle$. Find all subgroups of $G$.
We can easily spot the cyclic normal subgroup $C=\langle a \rangle = \{e,a,a^2,\dots,a^5\}$.
Now $C$ has four subgroups $\{e\}, \langle a^3 \rangle, \langle a^2 \rangle, C$ of order $1,2,3,6$.
Now apparently any other subgroups of $H$ of $D_{12}$ must intersect $C$ in one of these subgroups. Why is this? I dont want an explanation of how to find the next subgroups by some kind of brute force and intuition method.
The answer to your rather restricted question:
The intersection of two subgroups $A,B$ is a subgroup of both $A$ and $B$. In particular it must be one of the subgroups of $A$, so the intersection of any subgroup with $\langle a\rangle$ must be on the exhaustive list of subgroups of this group that you have enumerated.