Let $K=\Bbb Q(\alpha)$ be the field generated by the one root $\alpha\in\Bbb C$ specified in the OP.
Here is a pedestrian proof for $\sqrt 2\not \in K$. Assume for this that we have a tower of fields $\Bbb Q\subset \Bbb Q[\sqrt 2]=k\subset \Bbb Q[\alpha]=K$.
(So $K:k$ is an extension of degree two, it is Galois.)
Then the polynomial $X^4-2X^2+2$, seen as a polynomial in $k[X]$ splits as
$$
X^4-2X^2+2
=\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big)\ ,
$$
with $a,b,c,d\in\Bbb Q$.
The factor with the root $\alpha$ (and its $(K:k)$-conjugate) was only shown.
(The other factor does not have the root $\alpha$, so it is a different factor.)
Consier the above relation as a product of polynomials in $k[X]$. Then we also have its conjugate polynomial as a factor. So from
$$
\begin{aligned}
X^4-2X^2+2
&
=\Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)\cdot\Big(\dots\Big)
\\
&
=\Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\cdot\overline{\Big(\dots\Big)}
\end{aligned}
$$
we get (second factor is different, unique factorization)
$$
\begin{aligned}
X^4-2X^2+2
=&\ \Big(X^2+(a+b\sqrt 2)X+(c+d\sqrt 2)\Big)
\\
\cdot&\ \Big(X^2+(a-b\sqrt 2)X+(c-d\sqrt 2)\Big)\ ,
\end{aligned}
$$
then we multiply, look at the term in $X^3$, get $a=0$, then at the term in $X^1$, get $d=0$, so our factorization is
$$
X^4-2X^2+2
=
\Big(X^2+b\sqrt 2 X + c\Big)
\Big(X^2-b\sqrt 2 X + c\Big)\ ,
$$
so $c^2=2$. Contradiction.
This negates (1) in the OP, so (2) is out of charge.
To have a quick structure description of the Galois closure $L$ of $K=\Bbb Q(\alpha)$, let us start a parallel construction. Let $u=\zeta_8$ be the (cyclotomic) unit with minimal polynomial
$$
X^4+1
$$
over $\Bbb Q$, and to fix ideas, we will fix it with the complex image $\frac 1{\sqrt 2}(1+i)$ in the first quadrant. Its conjugates are $u, u^3, u^5, u^7$, with complex images $\frac 1{\sqrt 2}(\pm 1\pm i)$. It is clear that
$$
M=\Bbb Q(u)=\Bbb Q(\zeta_8)=\Bbb Q(\sqrt{-1},\sqrt 2)
$$
is (the cyclotomic field of order $8$, degree $4$ over $\Bbb Q$,) Galois over $\Bbb Q$ with Galois morphisms $\phi_j$ determined by $u\to\phi_j u=u^j$. So $\phi_5$ corresponds to $u\to u^5=-u$, and $\phi_7$ to the complex conjugation $u\to u^7=\frac 1u=\bar u$, they commute, the structure of $\operatorname {Gal}(M:\Bbb Q)$ is $\Bbb Z/2\times \Bbb Z/2$.
($\sqrt2$ is in $M$: $\zeta_8^2=\zeta_4=i$, so $\Bbb Q[i]$ is a subfield, but then from $\zeta_8=(1+i)/\sqrt 2$ also $\Bbb Q[\sqrt 2]$ is a subfield, and by dimension reasons we have $M=
\Bbb Q[\zeta_8]=\Bbb Q[\sqrt{-1},\sqrt 2]$. Then $\phi_5$ maps $i=u^2\to (-u)^2=u^2=i$, and thus $\sqrt 2=(1+i)/\zeta_8\to -\sqrt 2$. And $\phi_7$, the complex conjugation, does the other job, $i\to -i$, $\sqrt 2\to \sqrt 2$.)
We have now the "hint", that may be useful below,
$$
X^4-2X^2+2 = (X^2-1)^2-(\sqrt{-1})^2=(X^2-(1+u^2))(X^2-(1-u^2))\ .
$$
So we consider $N=M(a)=\Bbb Q(u,a)$, the quadratic extension of $M$ that introduces $a=\alpha=\sqrt{1+u^2}=\sqrt{1+\sqrt{-1}}$ as a root of the first factor. All roots of $X^4-2X^2+2$ are then
$$
a,\ -a,\ u^3a,\ -u^3a\ .
$$
We have a Galois substitution $\Psi:N\to N$ (over $M$, thus also over $\Bbb Q$), determined algebraically by $\Psi:(u\to u,\ a\to-a)$, and we lift $\phi_j$ as Galois morphisms $\Phi_j$ of $N$ as follows:
$$
\begin{aligned}
\Phi_5:&\ u\to\phi_5u=u^5=-u\ ,& (X^2-(1+u^2))&\text{ stays, so we can take } &\ a&\to a\ ,\\
\Phi_7:&\ u\to\phi_7u=u^7=\frac 1u=\bar u\ ,& (X^2-(1+u^2))&\text{ changes, so chose } & a&\to -u^3a\ .\\
&\qquad\text{Recall also:}
\\
\Psi:&\ u\to u\ , && & a&\to -a\ .
\end{aligned}
$$
($\Phi_7$ is well defined, in the relation $a^2-(1+u^2)=0$ the LHS is moved to $(-u^3a)^2-(1+u^6)=u^6a^2-1-u^6=u^6(a^2-1) -1 = (-i)(1+i-1)-1=0$. (Use $X$ instead of $a$ for a pedant factorization.)
In particular
$$
\boxed a
\overset{\Phi_7}{\longrightarrow}
\boxed {-u^3a}
\overset{\Phi_7}{\longrightarrow}
- u^{21}\; u^3a=
\boxed{-a}
\overset{\Phi_7}{\longrightarrow}
\boxed {u^3a}
\ ,
$$
so $\Phi_7$ permutes as a rotation the vertices of the square with the vertices marked as $a,-u^3a,-a,u^3 a$ in this cyclic order. Then $\Psi$ is the reflection w.r.t. the "virtual center" of the square, and $\Phi_5$ acts as the transposition of $\pm u^3 a$. (Symmetry w.r.t. the "virtula line" through $a,-a$, which are invariated.)
So $\Phi_7$ generates a Galois subgroup $\cong\Bbb Z/4$ and $\Phi_5$ implements / acts as a "twist" of it, the relation
$$
\Phi_5\Phi_7\Phi_5=\Phi_7^{-1}
$$
can be tested either geometrically (in the virtual picture) or algebraically on the generators, e.g.
$$
\begin{aligned}
\left(
\frac1u\leftarrow
\frac1{-u}\leftarrow
-u\leftarrow
u\right)
&=
\left(
\frac1u\leftarrow
u\right)
\ ,
\\
(u^3 a\leftarrow -u^3a \leftarrow a \leftarrow a)
&=
(u^3 a\leftarrow a)\ .
\end{aligned}
$$
It is a good idea to investigate the situation using computer power. Here sage. The following code gives information on the given situation.
sage: R.<X> = PolynomialRing(QQ)
sage: K.<a> = NumberField( X^4 - 2*X^2 + 2 )
sage: L.<s> = K.galois_closure()
sage: K.is_galois()
False
sage: L.degree()
8
sage: RK.<Y> = PolynomialRing(K)
sage: RK
Univariate Polynomial Ring in Y over Number Field in a with defining polynomial X^4 - 2*X^2 + 2
sage: RL.<Z> = PolynomialRing(L)
sage: RL
Univariate Polynomial Ring in Z over Number Field in s with defining polynomial X^8 - 20*X^6 + 104*X^4 - 40*X^2 + 1156
sage: a.minpoly()
x^4 - 2*x^2 + 2
sage: s.minpoly()
x^8 - 20*x^6 + 104*x^4 - 40*x^2 + 1156
sage: factor( Y^2 - 2 )
Y^2 - 2
sage: # so sqrt(2) is not in K
sage: factor( Z^2 - 2 )
(Z - 1/328*s^6 + 15/328*s^4 + 3/82*s^2 - 155/164) * (Z + 1/328*s^6 - 15/328*s^4 - 3/82*s^2 + 155/164)
sage: # so sqrt(2) is in L
sage: G.<g> = L.galois_group()
sage: G.structure_description()
'D4'
