Intermediary Extensions of $\mathbb{K}=\mathbb{Q}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$. First, I must prove this extensions is Galois, by some algebra I proved that:
$\mathbb{Q}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=\mathbb{Q}(\sqrt{2},i\sqrt{2})=\mathbb{Q}(\sqrt{2},i)$.
And since this last extensions is the splitting field of the irreducible polynomial $X^2+1 \in \mathbb{Q}[x]$ it is Galois with degree $4$.
Then, by the Galois correspondece the only intermediary fields are those with order $2$ with $\mathbb{K}$.
explicitly $G=Gal(\mathbb{K},\mathbb{Q})=\{Id, \sigma_1,\sigma_2,\sigma_3\}$, such that 
$\sigma_1(i)=i$, $\sigma_1(\sqrt{2})=-\sqrt{2}$
$\sigma_2(i)=-i$, $\sigma_2(\sqrt{2})=\sqrt{2}$
and
$\sigma_3(i)=-i$, $\sigma_3(\sqrt{2})=-\sqrt{2}$
All the $\sigma's$ are of order 2.
By the tower rule we can see that $\mathbb{Q(i)}$ and $\mathbb{Q(\sqrt{2})}$
are extensions with degree $2$ with $\mathbb{K}$, since they are degree 2 with $\mathbb{Q}$.
I would like to say that those two are the only fields that i'm looking for but i can't prove what are they respective images under the Galois Correspondence Theorem, i.e, what normal groups of $G$ they represent.
 A: You have shown that $G=\{\operatorname{id},\sigma_1,\sigma_2,\sigma_3\}$ and that $\sigma_i^2=\operatorname{id}$ for all $i$. It follows that $G\cong(\Bbb{Z}/2\Bbb{Z})^2$. The intermediary extensions of $\Bbb{Q}\subset K$ then correspond bijectively to the normal subgroups of $G$. More precisely, each subgroup $H\subset G$ corresponds to the subfield of elements of $K$ that are fixed by each element of $H$. That is to say, $H$ corresponds to the subfield
$$L:=\{x\in K:\ (\forall\sigma\in H)(\sigma(x)=x)\}.$$
In this way the whole field $K$ corresponds to the trivial subgroup $\{\operatorname{id}\}$, and the subfield $\Bbb{Q}$ corresponds to the entire group $G$. This leaves precisely three subgroups of $G$; the subgroups $H_i:=\{\operatorname{id},\sigma_i\}$ for $i=1,2,3$. These are subgroups of index $2$, so they correspond to extensions of $\Bbb{Q}$ of degree $2$.
For $\sigma_1$ we have $\sigma_1(i)=i$, so the subfield $\Bbb{Q}(i)$ is invariant under $\sigma_1$. Because $[\Bbb{Q}(i):\Bbb{Q}]=2$ it follows that $\Bbb{Q}(i)$ is the subfield corresponding to $H_1$.
Similarly, for $\sigma_2$ we have $\sigma_2(\sqrt{2})=\sqrt{2}$, so the subfield $\Bbb{Q}(\sqrt{2})$ is invariant under $\sigma_2$.  Because $[\Bbb{Q}(\sqrt{2}):\Bbb{Q}]=2$ it follows that $\Bbb{Q}(\sqrt{2})$ is the subfield corresponding to $H_2$.
So what is the subfield corresponding to $H_3$?
