Quadratic subfield of $\mathbb{Q}(\zeta)$

It's not true. Consider $\Bbb Q[\zeta_{12}]$. It contains both $\Bbb Q[i]$ and $\Bbb Q[\zeta_3]$, each of which have degree $2$ over $\Bbb Q$.

When $\zeta_p$ is a primitive $p$-th root of unity for an odd prime $p$, the cyclotomic field $K_p=\mathbf Q(\zeta_p)$ admits only one quadratic subfield because $Gal(K_p/\mathbf Q)$ is cyclic (this remains true for $K_4$). But for $K_n$ with composite $n$, you know that $Gal(K_n/\mathbf Q)\cong (\mathbf Z/n)^*$, which is not cyclic in general. To catch the quadratic subfields of $K_n$, you have just to decompose $(\mathbf Z/n)^*$ into a direct product of cyclic components.