Let the roots be $a,b,c,d,e$ with $a,b $ non-real. Your subgroup is transitive and has at least one transposition $(a b)$. Since it is transitive you can conjugate your transposition by some element to get a transposition which moves $c$ (conjugation preserves the cycle structure), and likewise one which moves $d$ or $e$.
Case 1. If this is $(bc)$ or $(ac)$ you can find your 3-cycle as a product of transpositions.
Case 2. If it is $(cd)$ then you can use conjugation to find a transposition which moves $e$ and whichever one you get, you can find your 3-cycle by composing it with either $(ab)$ or $(cd)$.
This isn't too hard to do, but the below method requires a little knowledge of the cubic resolvent $h$ of a quartic polynomial $g$, namely that $\operatorname{Gal}(h) \leq \operatorname{Gal}(g)$ and that the discriminants of $g$ and $h$ (essentially) coincide. (I can expand about this some if it would be helpful.)
If a polynomial $g$ is irreducible (which is a necessary condition for $\operatorname{Gal}(g) \cong S_{\deg g}$, and which we thus henceforth assume), then its Galois group acts transitively. The only transitive subgroups of $S_4$ are (up to conjugacy) $S_4, A_4, D_8, \Bbb Z_4, \Bbb Z_2 \times \Bbb Z_2$. We'll use the fact that the only groups among these whose order is divisible by $6$ are $S_4$ and $A_4$.
At least when the character of the underlying field $\Bbb F$ is not $3$, we may for simplicity of the below formulae make a linear change of variables so that $g$ has zero coefficient in its $x^3$ term and write (after dividing by the leading coefficient)
$$g(x) = x^4 + p x^2 + q x + r,$$
its resolvent cubic is
$$h(x) = x^3 - 2 p x^2 + (p^2 - 4 r) x + q^2,$$
and the discriminants of $g$ and $h$ coincide (perhaps up to an overall nonzero multiplicative constant), and are
$$D = 16 p^4 r − 4 p^3 q^2 − 128 p^2 r^2 + 144 p q^2 r − 27 q^4 + 256 r^3.$$
If $h$ is irreducible and its discriminant $D$ is not a square, then (1) $\operatorname{Gal}(h) \leq G := \operatorname{Gal}(g)$ is $S_3$, so $G$ has order divisible by $6$ and hence by the above is $S_4$ or $A_4$, and (2) since $D$ (the discriminant of $g$) is not a square, $G \not\leq A_4$ and hence $G \cong S_4$. For $g$ and $h$ to be irreducible, they must have nonzero constant terms and hence $q, r \neq 0$. For $p = 0$, the above formulae simplify to
$$h(x) = x^3 - 4 r x + q^2$$ and $$D = - 27 q^4 + 256 r^3,$$ so to find an example we can search for $q, r$ for which $h$ is irreducible and $D$ is nonsquare. For $\Bbb F = \Bbb Q$ the simple choice $q = r = 1$ satisfies these criteria (the first by the Rational Root Test), so, we have for example, that $$\operatorname{Gal}(x^4 + x + 1) \cong S_4 .$$
See these notes for more details (using the same notation). Also, note that these sorts of examples are generic in that, in a sense that can be made precise, most irreducible polynomials of degree $n$ in $\Bbb Q[x]$ have Galois group $S_n$.
Best Answer
There are $A_4$-Galois extensions of $\Bbb Q$. Let $L$ be one of them. The group $A_4$ has a subgroup $H$ of order $2$. Let $K=L^H$ be its fixed field. The subgroup $H$ is not normal in $A_4$; therefore the Galois closure of $K$ is $L$. As we are in characteristic zero, then $K=\Bbb Q(a)$ for some $a$. The minimal polynomial of $a$ over $\Bbb Q$ has degree six, and its Galois group is $A_4$.