Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$

You're off to a good start. The splitting field $\Omega$ contains $\sqrt[4]{2}e^{\tfrac a8\pi i}$ for $a\in\{\pm3,\pm5\}$ and hence also $$\frac{\sqrt[4]{2}e^{\tfrac 58\pi i}}{\sqrt[4]{2}e^{\tfrac 38\pi i}}=e^{\tfrac14\pi i}=\zeta_8,$$ which shows that $\Bbb{Q}(\zeta_8)\subset\Omega$. Over this subfield we already have $$X^4+2X^2+2=(X^2-(-1+i))(X^2-(-1-i)),$$ where $-1-i=\zeta_8^2(-1+i)$. So if $\alpha\in\Omega$ is a root of $X^2-(-1+i)$ then $$(\zeta_8\alpha)^2-(-1-i)=\zeta_8^2(\alpha^2-(-1+i))=0,$$ i.e. $\zeta_8\alpha$ is a root of $X^2-(-1-i)$. This shows that $\Omega$ is the splitting field of $X^2-(-1+i)$ over $\Bbb{Q}(\zeta_8)$. It now suffices to check that $-1+i$ is not a square in $\Bbb{Q}(\zeta_8)$ to conclude that $[\Omega:\Bbb{Q}]=8$.

Edit: In response to the comment below; an excellent hands-on proof of the fact that $-1+i$ is not a square in $\Bbb{Q}(\zeta_8)$ has already been given in another answer. So here's a less constructive proof:

Suppose $-1+i$ is a square in $\Bbb{Q}(\zeta_8)$. Then it is a square in its ring of integers $\Bbb{Z}[\zeta_8]$, which is a unique factorization domain, and the factorization $$-1+i =-(1-\zeta_8)(1+\zeta_8) =\zeta_8^2\tfrac{1+\zeta_8}{1-\zeta_8}(1-\zeta_8)^2,$$ shows that the unit $\tfrac{1+\zeta_8}{1-\zeta_8}\in\Bbb{Z}[\zeta_8]$ is then also a square, where $\tfrac{1+\zeta_8}{1-\zeta_8}=\zeta_8(1+\zeta_8+\zeta_8^2)$.

Because $\Bbb{Z}[\zeta_8]$ is a UFD, every unit in $\Bbb{Z}[\zeta_8]$ is a cyclotomic unit, i.e. of the form $$\zeta_8^a\big(\tfrac{1-\zeta_8^3}{1-\zeta_8}\big)^b =\zeta_8^a(1+\zeta_8+\zeta_8^2)^b,$$ for unique $a\in\Bbb{Z}/8\Bbb{Z}$ and $b\in\Bbb{Z}$. This shows that $\zeta_8(1+\zeta_8+\zeta_8^2)$ is not a square.