A degree $3$ Galois extension $K$ of $\mathbb Q$ has to be totally real. You're arguments essentially prove that: if $\alpha$ is a primitive element for $K$, then it satisfies a cubic $f$, which also has at least one real root $\beta$. But then $K = \mathbb Q(\beta)$, hence all the roots of $f$ are real.
Another, more sophisticated way to phrase the same thing is that if $v$ were a complex place of $K$, then the decomposition group of $v$, which is a copy of $\mathrm{Gal}(\mathbb C/\mathbb R)$, would embed into the Galois group of $K/\mathbb Q$;
but a group of order three can't contain a subgroup of order two.
In general, a Galois extension of $\mathbb Q$ is either totally real or totally complex. (This is a particular case of the general fact that for any place of $\mathbb Q$, the places of a Galois extension $K$ of $\mathbb Q$ that lie above the
given place of $\mathbb Q$ are acted upon transitively by $\mathrm{Gal}(K/\mathbb Q).$
One could also see it in terms of primitive elements: if $K = \mathbb Q(\alpha)$,
and $\alpha$ has minimal polynomial $f$, then all roots of $f$ are expressible in terms of any one of them (because $K$ is Galois and $\alpha$ is a primitive element) and hence all of them are real iff one of them is real.)
However, when people speak of cubic fields, often they mean the Galois closure of a degree $3$ field, i.e. the extension $K$ of $\mathbb Q$ obtained by adjoining all the roots of a cubic (which is degree $3$ over $\mathbb Q$ if the discriminant is a square, and degree $6$ otherwise). Then $K$ will be either totally real or totally complex, depending on whether the discriminant of the cubic is positive or not,
or equivalently, depending on whether or not all the roots of the cubic are real.
