No. Let $\alpha$ be a cube root of $2$. Then $\mathbb{Q}(\alpha)$ is not a Galois extension. Indeed, note that $\mathbb{Q}(\alpha)$ can be embedded in $\mathbb{R}$, so how could it possibly be the splitting field of $X^3 - 2$?
In the general case, we define a Galois extension to be an algebraic field extension $F / K$ such that the fixed field of $\mathrm{Aut}_K(F)$, the group of $K$-automorphisms of $F$, is not $K$. Notice that $\mathbb{Q}(\alpha)$ has no non-trivial $\mathbb{Q}$-automorphisms: thus it cannot possibly be a Galois extension. For finite extensions, by Dedekind's theorem on the linear independence of $K$-automorphisms, it suffices to check that $[F : K] = | \mathrm{Aut}_K(F) |$.
If a finite extension is not Galois, then that can only be because there are not enough automorphisms: Dedekind's theorem mentioned above in fact guarantees that $[F : K] \ge | \mathrm{Aut}_K(F)|$. In the separable case, if $F / K$ does not have enough automorphisms, that essentially means that $F$ does not contain enough roots.