$\mathrm{Gal}(\mathbb{Q}({\zeta }_{12})\mathrm{/}\mathbb{Q})\cong (\mathbb{Z}\mathrm{/}12{)}^{\times }$ and its three quadratic subfields

The units modulo $12$ are $[1],[5],[7],[11]$, since these are exactly the residue classes in $\{1,\dots ,11\}$ coprime to $12$. Their squares are $[25]=[1]$, $[49]=[1]$ and $[121]=[1]$, so every nonidentity element has order two.

Therefore $(\mathbb{Z}\mathrm{/}12{)}^{\times }$ is the Klein four-group, with three order-two subgroups: $H_i=\{[1],[5]\},H_{-3}=\{[1],[7]\},H_3=\{[1],[11]\}\mathrm{.}$ By [L1] and [L2], each fixed field has degree $2$ over $\mathbb{Q}$.

The subgroup $H_i$ fixes $i={\zeta }^3$, because ${\sigma }_5({\zeta }^3)={\zeta }^{15}={\zeta }^3$. Since $i\mathrm{\notin }\mathbb{Q}$ and the fixed field has degree $2$ by step 2.1, that fixed field is $\mathbb{Q}(i)$.

The subgroup $H_{-3}$ fixes $2{\zeta }^2-1$, because ${\sigma }_7({\zeta }^2)={\zeta }^{14}={\zeta }^2$; and $(2{\zeta }^2-1{)}^2=4{\zeta }^4-4{\zeta }^2+1=-3,$ since ${\zeta }^2$ is a root of $t^2-t+1$. So the fixed field contains $\sqrt{-3}$, and again step 2.1 makes it exactly $\mathbb{Q}(\sqrt{-3})$.

The subgroup $H_3$ fixes $\zeta +{\zeta }^{-1}$, because ${\sigma }_{11}$ is complex conjugation. Moreover $(\zeta +{\zeta }^{-1}{)}^2={\zeta }^2+2+{\zeta }^{-2}=3,$ since ${\zeta }^2+{\zeta }^{-2}=1$. So the fixed field contains $\sqrt{3}$, and step 2.1 makes it exactly $\mathbb{Q}(\sqrt{3})$.

The three order-two subgroups of step 2.1 therefore yield the three quadratic intermediate fields $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{3})$, and there are no others because [L2] gives a bijection between subgroups and intermediate fields.