No group has $\mathbf{N}_5$ as its lattice of subgroups

Let the group be $G$.

$B$ and $C$ cannot be conjugate as $B$ has a proper over-group and $C$ does not. So each is normal, and so we have a subgroup $BC$ of order $|B||C|$ containing $B$ and $C$. From the lattice $BC=G$.

$A$ and $C$ cannot be conjugate as $B$ has a non-trivial subgroup and $C$ does not. So $A$ is also normal, and so we have a subgroup $AC$ of order $|A||C|$ containing $A$ and $C$. From the lattice $AC=G$.

Hence $|G|=|A||C|=|B||C|$. But $|A|\not=|B|$ so we have a contradiction.

The group $A$ has a single non-trivial subgroup $B$, so it must be cyclic. More precisely, we can say that $|A|=p^2$ for some prime $p$ for otherwise it would have more non-trivial subgroups. Similarly, $C$ is cyclic and its order is a prime $q$.

If $C$ normalizes $B$, then $N_G(B)$ contains both $A$ and $C$, so $B\unlhd G$. In this case $BC$ is a subgroup, and your argument kicks in.

If $C$ does not normalize $B$, then $B$ has at least $q$ conjugate subgroups in $G$. They are all minimal (of order $p$) and all of them are contained in the corresponding conjugate of $A$, but there are no candidates for those conjugates of $A$ other than $A$ itself. So $A$ would need to contain $q$ non-trivial subgroups. A contradiction.

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I'm fairly sure that a simpler argument exists :-)