I've got that $x^4-4x^3+4x^2-3 =(x^2-2x+ \sqrt{3})(x^2-2x-\sqrt{3})$
The roots of the polynomials are:
$\alpha = 1+\sqrt{1-\sqrt{3}}$ $\quad$ $\alpha_1= 1-\sqrt{1-\sqrt{3}}$ $\quad$ $\beta= 1+\sqrt{1+\sqrt{3}}$ $\quad$ $\beta_1= 1-\sqrt{1+\sqrt{3}}$
By definition, the splitting field is $L=\mathbb{Q}(\alpha, \alpha_1, \beta, \beta_1)= \mathbb{Q}(\alpha, \beta) $
It's easy to show that $\alpha_1 \in\mathbb{Q}(\alpha)$ and that $\beta_1\in\mathbb{Q}(\beta)$.
Now my doubt is how to prove that $\beta \notin \mathbb{Q}(\alpha)$ And therefore the degree of the extension $L|\mathbb{Q}$ will be 8.
It's easier to prove that $\alpha\notin\mathbb{Q}(\beta)$, because this field is a subset of $\mathbb{R}$, but $\alpha\notin\mathbb{R}$.
The factorization you found over $\mathbb{R}$ shows the polynomial is irreducible over $\mathbb{Q}$, so $\beta$ has degree $4$.
We now need to compute the degree of $\alpha$ over $\mathbb{Q}(\beta)$. Since $\alpha$ is a root of $x^2-2x+\sqrt{3}$ which has coefficients in $\mathbb{Q}(\beta)$, we have that its degree is $2$.