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.