FALSE: every irreducible quintic over $\mathbb{Q}$ is insoluble by radicals

Eisenstein at the prime $2$ shows that $q(x)=x^5-2$ is irreducible over $\mathbb{Q}$, so it is an irreducible quintic.

By [L2], choose $s$ with $s^2=5$, choose $\beta$ over $\mathbb{Q}(s)$ with ${\beta }^2=\frac{-5-s}{2},$ and choose $\alpha$ with ${\alpha }^5=2$. Put $u:=\frac{-1+s}{2}\text{and}\zeta :=\frac{u+\beta }{2}\mathrm{.}$ Then $u^2+u-1=0$, ${\zeta }^2-u\zeta +1=0$, and therefore $\zeta +{\zeta }^{-1}=u\mathrm{.}$ Using $u^2+u-1=0$, this implies ${\zeta }^4+{\zeta }^3+{\zeta }^2+\zeta +1=0.$ Since $\zeta \mathrm{\ne }1$, we get ${\zeta }^5=1$, so $\zeta$ is a primitive fifth root of unity. Thus every root of $q$ has the form $\alpha {\zeta }^k$ with $0\le k<5$, and all of them lie in $L:=\mathbb{Q}(s,\beta ,\alpha )\mathrm{.}$

The field tower $\mathbb{Q}\subseteq \mathbb{Q}(s)\subseteq \mathbb{Q}(s,\beta )\subseteq \mathbb{Q}(s,\beta ,\alpha )=L$ adjoins successively a square root of $5$, a square root of $(-5-s)\mathrm{/}2$, and a fifth root of $2$. So $L\mathrm{/}\mathbb{Q}$ is a radical extension. Step 1.2 shows that the splitting field of $q$ is contained in $L$, and [L3] therefore makes $q$ solvable by radicals.

Hence $x^5-2$ is an irreducible quintic over $\mathbb{Q}$ that is solvable by radicals, so the universal claim is false.