Convert the question to a problem in permutation groups.
Let $F$ be the splitting field of
$$f(x)=(x-\theta_1)(x-\theta_2)(x-\theta_3)(x-\theta_4)$$
with $\theta=\theta_1$.
We were given that the Galois group realizes all the 24 permutations of the roots $\theta_i,i=1,2,3,4.$ Therefore
$$
\operatorname{Gal}(F/\mathbb{Q}(\theta))=\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})
$$
contains automorphisms realizing all the six permutations of the other roots.
Galois correspondence then means that the claim is equivalent to:
There are no subgroups $H$ properly between $\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})$ and $\operatorname{Sym}(\{\theta_1,\theta_2,\theta_3,\theta_4\})$.
In other words, this is equivalent to proving that the obvious copy of $S_3$ inside $S_4$ is a maximal subgroup. Have you seen that? If not, can you prove it?
As you already know, the elements of the Galois group are exactly: $\newcommand{\Q}{\mathbb{Q}} w \mapsto w, w \mapsto w^2, w \mapsto w^4, w \mapsto w^5, w \mapsto w^7, w \mapsto w^8$. $w \mapsto w^2$ is a generator, and we have
$$
w \mapsto w^2 \mapsto w^4 \mapsto w^8 \mapsto w^7 \mapsto w^5 \mapsto w.
$$
Thus we are looking for the fixed fields of $w \mapsto w^4$ and of $w \mapsto w^8 = w^{-1}$.
The fixed field of $\boldsymbol{\sigma: w \mapsto w^{-1}}$
As you have observed, $\alpha = w + w^{-1}$ is fixed by $\sigma$ and satisfies
$$
\alpha^3
= w^3 + w^{-3} + w^1 + w^{-1}
= (w^6 + 1)/(w^3) + \alpha
= \alpha - 1
$$
Since $\alpha^3 - \alpha + 1$ is irreducible over $\Q$, the desired fixed field is $\Q(\alpha) = \Q(w + w^{-1})$.
The fixed field of $\boldsymbol{\tau: w \mapsto w^4}$
$\beta = w^3$ satisfies $\beta^2 + \beta + 1$, which is irreducible over $\Q$ of degree $2$. So the desired fixed field is $\Q(\beta) = \Q(w^3)$.
A Remark
The generators for a subfield can end up being really simple: in our case $w + w^{-1}$ and $w^3$. But how do you find them?
The first thing to do is try looking for a single term or two terms added together that is fixed under the automorphism in question. In our case, $w^3$ is a single term so you would have found it quickly; $w + w^{-1}$ would not take much longer.
Alternatively, expecially if the first method fails, you can write an arbitrary element $z \in \Q(w) / \Q$ as $z = a w^5 + b w^4 + c w^3 + d w^2 + e w + f$, manually set $\sigma(z) = z$ and then solve for equations in $a, b, c, d, e, f$. You will use the minimal polynomial for $w$ (or in general, the minimal polynomial for a generator of whatever extension you are considering) to simplify powers of $w$ above $w^5$, and then equate each coefficient $w^0, w^1, \ldots, w^5$. This will give you a set of things fixed by the autorphism, and picking one such list of $a, b, c, d, e, f$ chances are good you will get a generator of the desired field.
A Second Remark
This method should always work, though it could potentially be more tedious.
Say we have an extension $F(\alpha) / F$ of degree $n$ with Galois group $G$.
First, enumerate all the automorphisms of the extension by casework on where they send $\alpha$, and use them
to classify the Galois group $G$.
Then enumerate the subgroups, and for each subgroup $H$, look for elements $z \in F(\alpha)$
that are fixed by $H$.
Once you have an element $z$ with $[F(z) : F] = [G : H] = n / |H|$, you know that
$z$ generates the corresponding extension.
This works for any $H$; $H$ need not be normal in $G$.
For a Galois extension $K / F$ with group $G$ the size of a subgroup $H$
always equals the degree $[K : E]$, where $E$ is the fixed field of $H$.
In other words, the number of automorphisms of $K / F$ fixing $E$ equals the degree of $K / E$.
Equivalently, $\boldsymbol{K/E}$ is Galois for any $H$.
What normality of $H$ corresponds to is that $\boldsymbol{E / F}$ is Galois, i.e. that the number of automorphisms
of $E / F$ fixing $F$ equals the degree $[E : F]$.
In this case the Galois group of $E / F$ is the quotient group $G / H$.
A good example is the extension $K / \Q$, where $K$ is the splitting field of $x^3 - 2$.
If $\omega$ is a cube root of unity, the subextensions are generated by
$\omega, \sqrt[3]{2}$, $\omega \sqrt[3]{2}$, and $\omega^2 \sqrt[3]{2}$,
and have degree 2, 3, 3, and 3, respectively over $\Q$.
The Galois group is $S_3$ and the corresponding subgroups are
of index 2 (normal), index 3 (not normal), index 3 (not normal), and index 3 (not normal), respectively.
So the index of the subgroup always equals the degree of the extension.
Normality doesn't factor in unless we consider the automorphism group of a subextension rather than
the subgroup fixing it.
Best Answer
It looks like this problem is leading up to the fundamental theorem of Galois theory, which states that there is a one-to-one correspondence between the subfields of a Galois extension and subgroups of its Galois group. So our answer will be, of course, that no such Galois extension exists (which is why it's difficult to find an example).
First, note that if $K$ is a Galois extension of degree $4$ over $\mathbb{Q}$, then it must be the splitting field of a quartic polynomial. Either that polynomial is irreducible, or it can break into a product of irreducible quadratics. Of course, you'll want convince yourself of these cases and that the cases are indeed exhaustive.
If the polynomial breaks into a product of two irreducible quadratics, then I claim that it's rather easy to show that $K$ must have a subfield.
On the other hand, if the polynomial is irreducible, then $K = \mathbb{Q}[\alpha]$, where $\alpha$ is one of its roots. Since the degree of the Galois extension is $4$, we also know the Galois group has order $4$. As such, the Galois group must have an element $\phi$ of order $2$ that sends $\alpha \mapsto \beta$, where $\beta$ is another root of the polynomial.
Claim: $\mathbb{Q}[\alpha\beta] \subsetneq \mathbb{Q}[\alpha]$
To show this, suppose for contradiction they were actually the same field. Then both would be of degree $4$ over $\mathbb{Q}$, and the former would have a basis $\{1, \alpha\beta, (\alpha\beta)^2, (\alpha\beta)^3\}$. Because $\alpha$ would be an element of both fields, we could write:
$$\alpha = c_1 + c_2(\alpha\beta) + c_3(\alpha\beta)^2 + c_4(\alpha\beta)^3$$
What happens when the automorphism $\phi$ acts on both sides of this expression? What can we conclude?