I've been wondering about this for a while, is it the case that if $K$ is a finite Galois extension of $\mathbb{Q}_p$, then there exists some field $L$ with $$\mathbb{Q} \subset L \subset K,$$ such that $[L:\mathbb{Q}]<\infty$ and $L\cdot \mathbb{Q}_p = K$.
I can see that this is true for $K$ unramified, because we can just attach the appropriate roots of unity to $\mathbb{Q}$ so it remains the case where $K$ totally ramifies. I guess it would be enough to show that there exists some $\alpha \in K$ which is algebraic over $\mathbb{Q}$ but not an element of $\mathbb{Q}$. I have tried to use Hensel's lemma but that fails spectacularly because the uniformizer is the root of some Eisenstein polynomial which reduces to be $x^n$.
Yes, this is Hensel-Krasner lemma. Given a square-free polynomial $f\in \Bbb{Z}_p[x]_{monic}$, there is some $N$ such that for $n>N$, if $f(a_n) = 0\bmod (\pi^n), a_n\in O_K$ then $f$ has a root in $O_K$. The proof is the same as when $f$ is separable $\bmod p$ :
$$f(a_n+b)= f(a_n)+b f'(a_n)+O(b^2/p)$$ Normalize such that $v(\pi)=1$. For $N$ large enough we have $v(f'(a_n))\le m$.
With $b=-f(a_n)/f'(a_n)$, $a_{n+1}=a_n+b, 2(n-m)-v(p) > n$ ie. $N > 2m+v(p)$ we get that
$v(f(a_{n+1}))>v(f(a_n))$ and hence $\lim_{n\to \infty} f(a_n)=0$, by compactness $a_n$ has a subsequence converging in $O_K$, to a root of $f$.
Next $K= \Bbb{Q}_p(\pi+\zeta_{q-1})\cong \Bbb{Q}_p[x]/(g(x))$, take $f\in \Bbb{Z}[x]$ of same degree approximating $g$ with enough precision so that its $m$ is the same as $g$ and $f(\pi+\zeta_{q-1})=0\bmod (\pi^{2m+v(p)+2})$.
Take $a_{N+2} = \pi+\zeta_{q-1}$.
$f$ will have a root $a\in K$, which is $\equiv \pi+\zeta_{q-1}\bmod (\pi^2)$, whence $\Bbb{Q}_p(a)$ has the same residue field and uniformizer as $K$ ie. $K=\Bbb{Q}_p(a)$.