Prove that $[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}] = 8$.

If you aim only to show that $[\mathbf Q (\sqrt{4 \pm \sqrt 5}):\mathbf Q]=8$, it's much quicker to introduce the quadratic field $k=\mathbf Q(\sqrt 5)$ and consider the quadratic extensions $k(\sqrt{4 \pm \sqrt 5})/k$. Because $\mathbf Q(\sqrt{4 \pm \sqrt 5})$ contains $\sqrt 5$, it's obvious that $k(\sqrt{4 \pm \sqrt 5})=\mathbf Q(\sqrt{4 \pm \sqrt 5})$. Now, it's well known and straightforward that $k(\sqrt x)=k(\sqrt y)$ iff $xy\in {k^*}^2$ (this is a particular case of Kummer's theory). Here $xy=16-5=11$ cannot live in $k^*$ : this can be checked "by hand", or by using again the fact that $\mathbf Q(\sqrt 5)=\mathbf Q(\sqrt 11)$ iff $55$ is a square in $\mathbf Q^*$, which is impossible. So the two quadratic extensions $k(\sqrt{4 \pm \sqrt 5})/k$ are distinct, their compositum is biquadratic over $k$, and you're done.

Let $P= x^4- 8 x^2 + 11$ and $\pm\theta, \pm\theta'$ the roots of $P$ in $\mathbb R$, $\theta, \theta'$ positive.

Since $$ \theta \cdot \theta' =\sqrt{11} $$ then $$ \mathbb Q(\theta, \theta')=\mathbb Q(\theta,\sqrt{11}) $$ and $[\mathbb Q(\theta, \sqrt{11}):\mathbb Q(\theta)]=:n$ is either $1$ or $2$. Let's see that it is $2$.

A possible strategy using Galois theory: Assume that $n=1$, then $[\mathbb Q(\theta, \sqrt{11}):\mathbb Q(\sqrt{11}]=2$ so the minimal polynomial of $\theta$ over $\mathbb Q(\sqrt{11})$ is degree $2$. This is a divisor of $P$ thus it is one of the following. $$ \begin{array}{c} P_1=(x-\theta)(x+\theta)\\ P_2=(x-\theta)(x-\theta')\\ P_3=(x-\theta)(x+\theta') \end{array} $$ It is easy to see that $P_1\notin \mathbb Q(\sqrt{11})[x]$. Indeed $P_1(x) = x^2 - 4 - \sqrt{5}$ and $\sqrt{5}\notin \mathbb Q(\sqrt{11})$. It is slightly harder to prove that $P_i$ is not in $\mathbb Q(\sqrt{11})[x]$ for every $i=2,3$. This leads to a contradiction.

Using $29$-adics: $P$ has a simple root $14$ in $\mathbb F_{29}$. By Hensel's Lemma, $\mathbb Q(\theta)$ admits an embedding in $\mathbb Q_{29}$, field of $29$-adic numbers.

($29$ is the smallest prime $q$ such that $P$ has a simple root in $\mathbb F_{q}$).

If $n=1$, that is if $\sqrt{11}\in \mathbb Q(\theta)$, then $\mathbb Q_{29}$ contains a root of $X^2 -11$. Hence $11$ is a square in $\mathbb F_{29}$. This is a contradiction.

Equivalently, since $P$ factors in irreducible factors as $$ (X+ 14) (X+ 15)(X^2 +14) $$ in $\mathbb F_{29}$ then $\mathbb Q_{29}$ only contains $2$ out of $4$ roots of $P$. This is a $p$-adic version of the following argument over $\mathbb C$: $\mathbb Q(\sqrt{1+\sqrt{7}})/\mathbb Q$ is not Galois since its minimal polynomial has real and complex non-real roots.

Edit: Following OP's strategy assume that $\theta \in \mathbb Q(\sqrt{11},\sqrt{5})$. Since $G=\mathbb Q(\sqrt{11},\sqrt{5})/\mathbb Q$ is Galois and $P$ is irreducible then $G$ acts transitively on $S=\{\pm\theta,\pm\theta'\}$. In particular $$ S=G(\alpha) :=\{ g(\theta):g\in G\} $$ for every $\alpha \in S$. Hence $S=\{\theta,\sigma(\theta),\tau(\theta),\sigma\tau(\theta)\}$. But by OP's argument $\tau(\theta) =\pm \theta$, hence $\tau(\theta)=-\theta$. Similarly one prove that $\tau(\theta') = -\theta'$. Hence $$ \sqrt{11}=\theta \theta' = \tau(\theta\theta') = \tau(\sqrt{11}) = -\sqrt{11}. $$