An irreducible polynomial in $\mathbb Q[x]$ that has all zeros with multiplicity $2$

If $F$ is a field of characteristic zero and $f\in F[x]$ is irreducible then $f$ is separable. The easiest way to see this is define the formal derivative $f'$. (which looks exactly like the derivative from calculus but it is formal, there is no convergence here). It is easy to see that $\alpha$ is a root of $f$ with multiplicity higher than $1$ if and only if $f'(\alpha)=0$. It follows that $f$ is separable if and only if $\gcd(f,f')=1$.

Now, if $f$ is irreducible then $\gcd(f,f')\in\{1,f\}$. Since we are working in a field of characteristic zero we have $deg(f')=def(f)-1$ and hence the $\gcd$ can't be $f$. So it must be $1$ and hence $f$ is separable. This is actually a way to prove that any algebraic extension of a field of characteristic zero is separable.

So the point is you can't find an irreducible polynomial in $\mathbb{Q}[x]$ which has roots with multiplicity $2$.

There is none; only reducible ones.

If there were an irreducible one, $f$, we could adjoin a root, and we'd have a finite extension of $\Bbb Q $ (of $\operatorname {deg}f$), namely $\Bbb Q[x]/(f)$, which wasn't separable.

This violates theorem $51.13$, since $\operatorname {char}\Bbb Q=0$.