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).