Let $f(X)\in \Bbb Q[X]$ be a polynomial with $\deg f(X)=4$ and $E$ be its splitting field over $\Bbb Q$.
We would like to show that the Galois group $\mathrm{Aut}(E/\Bbb Q)$ of $f(X)$ is solvable and to deduce that the polynomial equations of degree $4$ over $\Bbb Q$ are solvable by radicals.
My attempt. Since $\deg f(X)=4$, it has at most $4$ roots over $\Bbb Q$. Let $\alpha_1,\dots,\alpha_k\in E$, with $0\leq k \leq 4$ be the distinct root that at outside $\Bbb Q$. Then $E:=\Bbb Q(\alpha_1,\dots,\alpha_k)$. Now we look at the group $\mathrm{Aut}(\Bbb Q(\alpha_1,\dots,\alpha_k)/\Bbb Q)$. Then, this group is completely determined by the effect of an automorphism to the elements $\alpha_1,\dots,\alpha_k$. Since these elements via a $\Bbb Q$-automorphism are again roots of $f(X)$, we obtain that $|\mathrm{Aut}(\Bbb Q(\alpha_1,\dots,\alpha_k)/\Bbb Q)|\leq 4!$. But we know that $\mathrm{Aut}(\Bbb Q(\alpha_1,\dots,\alpha_k)/\Bbb Q)$ an be embedded to $S_4$. The latter is solvable, so the Galois group of the polynomial $\mathrm{Aut}(E/\Bbb Q)$ is solvable. Therefore, by the Galois' Theorem for solvability (since $\mathrm{Char}(\Bbb Q)=0$), the polynomial equation $f(X)=0$ is solvable by radicals.
Question: Is this proof, as written, completely correct and complete?
Any other comment is very welcome.