- $\Bbb Q\left(\sqrt{5+\sqrt{21}} + \sqrt{5-\sqrt{21}}\right)$
- $\Bbb Q\left(\sqrt{5+\sqrt{21}} - \sqrt{5-\sqrt{21}}\right)$
For an involutive automorphism $\phi$, for all elements $s$, we have that $s+\phi(s)$ and $s\cdot \phi(s)$ are always fixed, so they have a chance to generate the subfield..
Update: interestingly,
$$ \begin{align} \left(\sqrt{5+\sqrt{21}} + \sqrt{5-\sqrt{21}}\right)^2 &= 14 \\
\left(\sqrt{5+\sqrt{21}} - \sqrt{5-\sqrt{21}}\right)^2 &= 6.
\end{align} $$
Hmm.. In other words, the following 3 intermediate fields we have:
$$\Bbb Q(\sqrt 6),\ \Bbb Q(\sqrt{14}),\ \Bbb Q(\sqrt{21})$$
There is a general method to deal with these sorts of extensions. Suppose we have $F$ a field of characteristic not equal to $2$ and suppose we adjoin $\alpha = \sqrt{a + b\sqrt{c}}$ Suppose that $b\neq0$, $a^2 - b^2c = h^2$ for some $h\in F$. Let $f(x)$ be the minimal polynomial of $\sqrt{a + b\sqrt{c}}$ (put some nice conditions on $a,b,c$ so that $F(\alpha)$ is an extension of degree $4$).
Let us call $\alpha_1 = \alpha$ , $\alpha_2 = \sqrt{a - b\sqrt{c}}$, $-\alpha_1= \alpha_3$ and $-\alpha_2 = \alpha_4$. We choose the following presentation for $V_4$ (the Klein 4- group isomorphic to $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$): $$V_4 = \{e, (12)(34), (13)(24),(14)(23)\}.$$ Note that $(14)(23) = \big((13)(24)\big)\big((12)(34)\big)$. Then we can see that each cycle in $V_4$ acts on the roots according to the numbering convention: For example the cycle $(12)(34)$ sends $\alpha_1 \mapsto \alpha_2$, $\alpha_3\mapsto \alpha_4$. Now for the sake of simplicity let us call
$$\theta = (12)(34), \hspace{5mm} \gamma = (13)(24), \hspace{5mm} \gamma\theta = (14)(23).$$
Then it is easy to see that $\theta(\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2, \gamma\theta(\alpha_1 - \alpha_2) = \alpha_1 - \alpha_2$. Now because $\gamma(\alpha_1) = \alpha_3$, squaring both sides we see that $\gamma(\sqrt{c} ) = \sqrt{c}$. It follows from these observations that
$$F(\alpha_1 + \alpha_2) \subset E^{\langle \theta \rangle}, \hspace{5mm} F(\sqrt{c}) \subset E^{\langle \gamma \rangle}, \hspace{5mm} F(\alpha_1 - \alpha_2) \subset E^{\langle \gamma\theta\rangle }.$$
By computing minimal polynomials/degrees we get that these subset inclusions are actually equalities and so you are done.
Edit: As Jyrki points out, the only presentation for $V_4$ that we can choose is the one given in my answer as that presentation is the only one which gives a transitive subgroup of $S_4$.
Best Answer
Let $\alpha=\sqrt{(1+\sqrt{5})/2}$ and $\beta=\sqrt{(-1+\sqrt{5})/2}i$. So the roots of $f(x)=x^4-x^2-1$ are $\pm \alpha$ and $\pm \beta$.
You might want to start a building project. Build a few towers: $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{5}) \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\alpha,i)$. You also have a tower $\mathbb{Q} \subseteq \mathbb{Q}(i) \subseteq \mathbb{Q}(i,\sqrt{5}) \subseteq \mathbb{Q}(\alpha,i)$. You can also build a tower by first adjoining $\sqrt{5}i$.
Recall that if $D_4 = \{1,r,r^2,r^3,s,rs,r^2s,r^3s\}$, then the subgroups are $D_4$, $\{1,r,r^2,r^3\}$, $\{1,r^2,s,r^2s\}$, $\{1,r^2,rs,r^3s\}$, $\{1,r^2\}$, $\{1,s\}$, $\{1,rs\}$, $\{1,r^2s\}$, $\{1,r^3s\}$, and $\{1\}$.
Notice that you can identify several of these already ($\mathbb{Q}(i,\sqrt{5})$ has the Klein 4-group as a Galois group. The degree of $\mathbb{Q}(\alpha,i)$ over $\mathbb{Q}(i,\sqrt{5})$ is 2 so this subfield is fixed by one of the subgroups of order 2. It's a splitting field of $(x^2+1)(x^2-5)$ so it is Galois itself, thus it's fixed by $\{1,r^2\}$ (the only normal subgroup of order 2).
You should be able to fill in most everything else by using the lattice correspondence and intersecting/joining subfields.
I hope this helps you get started. :)
EDIT: Now that you've had time to fight with your example for a while. I should note that this exact splitting field is dealt with in Keith Conrad's blurb/handout Galois correspondence examples. Specifically it's pages 7-10 (example #5).