To find the Galois group, I first begun by finding the splitting field:
I substituted $u=x^2$ and found the roots of the polynomial to be $\sqrt{3}\cdot e^{i\frac{\pi(2k+1)}{4}}$ for $k=\{0,1,2,3\}$. So the splitting field is generated by a single root, when $k=0$.
Applying Eisenstein to $(x+1)^4+9=x^4+4x^3+6x^2+4x+10$, with $p=2$, we see that it is irreducible. Hence the extension is of degree $4$. Now, the Galois group must have order $4$ too, and since the polynomial is separable and irreducible, the Galois group must be a transitive subgroup of $S_4$. There are two transitive subgroups of $S_4$, one is isomorphic to $Z_2\times Z_2$, and the other to $Z_4$.
To choose between those two, I computed the discriminant which was $4^4\cdot 3^6$ which is a perfect square in $\mathbb{Q}$. Hence the Galois group is a subgroup of $A_4$, and it must thus be isomorphic to $Z_2\times Z_2$.
Is my proof correct? Is there a better way to go through this problem? I would have preferred to see how the roots interact, but I couldn't manage to do it.
I don´t see anything wrong with what you've done, but you say you're interested in a different approach.
Recall that the primitive $8$-th root of unity is $\frac{1+i}{\sqrt{2}}$, so that your four roots are $\sqrt{\frac{3}{2}}(\pm 1 \pm i)$. A little manipulation shows that the splitting field is just $\mathbb{Q}[\sqrt{\frac{3}{2}}, i]$. The automorphisms must now swap or fix $\pm \sqrt{\frac{3}{2}}$ and swap or fix $\pm i$, so the group is indeed the fours-group. The action on the four roots of $X^4+9$ is easy to compute.