The definition I have for a normal closure is $E \subset G$ is a normal closure if any normal extension of $F$ containing $E$ contains a subfield F-isomorphic to $G$
Existence: Take$G$ to be the intersection of all normal extensions of $E$ (already proved there exists at least one) then this will be a normal extension and obviously there will be a subgroup that is F-isomorphic to $G$ since $G$ is contained in every normal extension. Am I missing any details?
Uniqueness (up to isomorphism): Here is where I am struggling. I assume uniqueness follows somehow from the uniqueness of a splitting field, but it may also be necessary to suppose that there are 2 and take an element the is in one and not the other and arrive at a contradiction somehow, but I don't see how.
Example: $F = \mathbb{Q},E = \mathbb{Q}(\sqrt[3]2), G$ a normal closure of $E \supset F$. Describe Gal($G/F$). $[E:F] = 3$ because {$1, \sqrt[3]2, \sqrt[3]2^2$} forms a basis. So the normal closure would be the splitting field. Is that $\mathbb{Q}(\sqrt[3]2, \sqrt[3]4)$? I'm having trouble finding it. Then do I just find [G:F] and then find a group of that order and that is isomorphic to the Galois group
Existence: Looks pretty good.
Uniqueness: Suppose there are two, $G_1$ and $G_2$. You don't need to consider elements in one and not the other and derive a contradiction; an easier path would be to try to prove $G_1$ contains $G_2$ and vice versa, so they must be equal. Can you do this from the definition of normal closure?
Example: Since $\sqrt[3]{4} = \sqrt[3]{2}^2$, $\mathbb Q(\sqrt[3]2,\sqrt[3]4)$ is just the same as $\mathbb Q(\sqrt[3]2)$. So it can't be the normal closure unless $\mathbb Q(\sqrt[3]2)$ is already normal over $\mathbb Q$. Is it? If not, why not? This will help you find the normal closure.