I would like some feedback on my proof of the fact stated in the question.
First, we note that $ L = \mathbb{Q}(\zeta_5) $ is the splitting field of $ X^5 - 1 $, so it is Galois. Furthermore, we know that $ \mathrm{Gal}(\mathbb{Q}(\zeta_5)) \cong C_{4} $. By the fundamental theorem of Galois theory, the subfields of $ L $ of degree 2 are in one-to-one correspondence with the subgroups of the Galois group which have $ 4/2 = 2 $ elements. However, $ C_4 $ has only one subgroup of order 2, which is $ \{e, g^2\} $ where $ g $ is an arbitrary generator. We conclude that $ L $ has only one subfield of degree 2, and it is fixed by the aforementioned subgroup of the Galois group.
Now, the Galois group is comprised of automorphisms $\sigma_i: \zeta_5 \to \zeta_5^i $ for $1 \leq i \leq 4 $, and this group has only one element of order 2, which is $\sigma_4$. We conclude that the intermediate field $ F $ is the one fixed by the cyclic subgroup generated by $ \sigma_4 $. Now, we consider the polynomial $ P(X) = (X - \zeta_5)(X - \zeta_5^4) $. Again, by the fundamental theorem, $\textrm{Gal}(L/F) = \{\sigma_1, \sigma_4\}$. We see that the entire Galois group fixes $P$, so $P$ has coefficients in $F$. Defining $ A = \zeta_5 + \zeta_5^4 $ which is negative of the coefficient of $X$ in $P$, we see that
$ A^2 + A - 1 = \zeta_5^2 + \zeta_5^3 + \zeta_5 + \zeta_5^4 + 1 = 0 $
and therefore, by the quadratic formula, $$ A = \frac{-1 \pm \sqrt{5}}{2} $$ and $ \sqrt{5} \in F $, implying that $ \mathbb{Q}(\sqrt{5}) = F $ as $ F $ has degree 2.
