Problem: Let $K$ be a field with char$(K) \neq 2$ and let $E/K$ be a field extension. Prove the following statements are equivalent:
(i) $E/K$ is a Galois extension of degree $4$ and $Gal(E/K) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
(ii) $E = K(\alpha, \beta)$ with $\alpha, \beta \in E \setminus K$, $\alpha \cdot \beta \notin K$ and $\alpha^2, \beta^2 \in K$.
Hint: Prove the following lemma first, and then use it:
Lemma: If Char$(K) \neq 2$ and $L/K$ is a field extension of degree $2$, then $L = K(\gamma)$ with $\gamma^2 \in K$.
Attempt:
Proof of the lemma: Suppose char$(K) \neq 2$ and let $K \subset L$ be a field extension of degree $2$. Let $\gamma \in L \setminus K$. We wish to show $L = K(\gamma)$. We have $$ 2 = [L: K] = [L: K(\gamma)] [K(\gamma) : K]. $$ Now $K(\gamma) \neq K$ because $\gamma \notin K$, and thus $[K(\gamma) : K] \neq 1$. It follows that $[L:K(\gamma) ] = 1$ because $2$ is prime. So $L = K(\gamma)$. I wish to show also that $\gamma^2 \in K$, but I don't know how. Where do I use that Char$(K) \neq 2$ for this?
Proof of theorem:
Suppose that we had proven the lemma.
I now wish to show $(i) \Rightarrow (ii)$. It is given that $Gal(E/K) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \cong V_4$ where $V_4$ is the Klein four group. Since $V_4$ is Abelian, every subgroup is normal. In particular, we know there exists a subgroup $H$ of order $2$. By the Galois correspondence, we know there must exist a normal (intermediate) field extension $L$ such that $$K \subset L \subset E$$ and such that $[L: K] = 2$. From the lemma, we have $L = K(\alpha)$ for some $\alpha \in L \setminus K$ and $\alpha^2 \in K$. Also, it is clear that $[E: L] = 2$. Again by the lemma, we may write that $E = L(\beta)$ with $\beta \in E \setminus L$ and $\beta^2 \in L$.
Now I wanted to claim that $E = K(\alpha, \beta)$. But I'm not sure how this follows from what I said previously, and also I have that $\beta^2 \in L$, while I need $\beta^2 \in K$. And the fact that $\alpha \cdot \beta \notin K$ is also not clear to me.
For the converse, $(ii) \Rightarrow (i)$ I didn't have a lot of inspiration. I wrote $$ [E: K] = [K(\alpha, \beta): K] = [K(\alpha)(\beta): K(\alpha] [K(\alpha):K]. $$
From this I still need to prove somehow that $K \subset E$ is a Galois extension of degree $4$. I think one can argue the isomorphism by claiming there are only two groups of order $4$, and then appealing to order of subgroups perhaps, so that it must be the Klein four group.
Help with finishing this problem is appreciated!
For the proof of the lemma, you realized that $K[\gamma]=L$, since its degree is $2$, $\gamma^2=a\gamma+b$, this implies that $(\gamma-{a\over 2})^2=c-{a^2\over 4}$, write $c=\gamma-{a\over 2}, K[c]=K[\gamma]$.
Proof of the theorem. $Gal(E/K)$ is the product of two groups $I,J$ which are isomorphic to $\mathbb{Z}/2$, as you have remarked, let $L^I$ the subfield of $L$ fixed by $I$, $[L^I:K]=2$, we can write $L^I=K(\alpha), L^J=K(\beta), \alpha^2,\beta^2\in K$. Remark that if $x\in L^I\cap L^J$, $x$ is fixed by $I$ and $J$, this implies that $x\in K$. You deduce that $[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha)][K(\alpha):K]$, $[K(\alpha,\beta):K(\alpha)]=1$ implies that $\beta\in K(\alpha)$ and $L^J\subset L^I$ contradiction, you deduce that $ [K(\alpha,\beta):K(\alpha)]=2$, and $K(\alpha,\beta)=L$.