Show that the Galois extension is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$ $E$ is the splitting field of $f(x)\in\mathbb{Q}[x]$ s.t. $[E:\mathbb{Q}] = 4$. $E$ has 2 distinct quadratic subextensions, $\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$. Prove that $Gal(E/\mathbb{Q}) \cong \mathbb{Z}_2\times \mathbb{Z}_2$ and that every quadratic subextension of $E$ is either $\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$ or $\mathbb{Q}(\sqrt{mn})$.
My attempt:
1) $E \cong \mathbb{Q}(\sqrt{m},\sqrt{n})$ and the minimal polynomial for $\sqrt{m},\sqrt{n}$ is $(x^2-m)(x^2-n)$ the roots of which are $\pm \sqrt{m},\pm \sqrt{n}$. Let $Gal(E/\mathbb{Q}) = \{e,\sigma,\tau,\sigma \tau\}$. Let $\sigma(\sqrt{m}) = -\sqrt{m}$. Now I need to show that $\sigma(\sqrt{n}) = \sqrt{n}$ to conclude that the order of $\sigma = 2$. I don't know how use to use the fact that $\sigma$ is a homomorphism to conclude that?
2) I can see that the only subextensions $K$ of $E$  such that $4 = [E:\mathbb{Q}] = [E:K][K:\mathbb{Q}]$ and $[E:K] = [K:\mathbb{Q}] = 2$ are $K=\mathbb{Q}(\sqrt{m}),\mathbb{Q}(\sqrt{n})$ or $\mathbb{Q}(\sqrt{mn})$ but how can I formally say that? Thanks!

For the first part, you could simply argue from the following facts :
(i) $\Bbb Z_2 \times \Bbb Z_2$ is the only group of order 4 with two distinct subgroups of index 2 and (ii) the Fundamental Theorem Of Galois Theory.
For the second part, again by the Fundamental Theorem Of Galois Theory, $E| \Bbb Q$ has 3 distinct quadratic subextenions and $\Bbb Q(\sqrt{m}), \Bbb Q(\sqrt{n})$ are two such (distinct), thus all you are left to show that $ \sqrt{mn} \notin\Bbb Q$ ( Since, $\sqrt{mn} \in \Bbb Q(\sqrt{m}) \implies \sqrt{n} \in \Bbb Q(\sqrt{m}) \implies \Bbb Q(\sqrt{n}) \subseteq \Bbb Q(\sqrt{m})$ contradicting that those two are distinct quadratic extensions) . But if $\sqrt{mn} \in \Bbb Q$ then  $\sqrt{mn} \in \Bbb Q(\sqrt{m}) \implies \sqrt{n} \in \Bbb Q(\sqrt{m})$ contradiction!
