Let $p$ be a prime, and let $K/\mathbb{Q}_p$ be a tamely ramified finite extension of degree $n$. Let $q$ be a prime factor of $n$ with $q\neq p$. Must there exist an intermediate extension $L$ (with $\mathbb{Q}_p\subset L\subset K$) of degree $q$ over $\mathbb{Q}_p$?

Some partial progress, following Hasse's "Number Theory", Chapter 16:

We can decompose our field extension as $\mathbb{Q}_p\subset K^{ur}\subset K$, where $K^{ur}$ is the maximal unramified subextension. Let $e=[K:K^{ur}]$ and $f=[K^{ur}:\mathbb{Q}_p]$. If $q$ divides $f$, we are done, since $K^{ur}/\mathbb{Q}_p$ is Galois with cyclic Galois group.

Otherwise, write $K^{ur}=\mathbb{Q}_p(\zeta)$ and $K=\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/e})$, where $\zeta$ is a primitive $(p^f-1)st$ root of unity, $u$ is a unit in the $p$-adic integers, and $0\leq r<\gcd(p^f-1,e)$. If $\gcd(p^f-1,e)=1$, then $K$ contains $\mathbb{Q}_p((up)^{1/e})$, and thus the intermediate field $\mathbb{Q}_p((up)^{1/q})$, which has degree $q$ over $\mathbb{Q}_p$.

We can also reduce to the case of $e=q$, by considering the subfield $\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/q})$. Since $q$ is prime and we dealt with the relatively prime case, we need only consider the case of $q$ dividing $p^f-1$. Note that if $q$ divides $p-1$, we have that $\mathbb{Q}_p(\zeta,(up\zeta^r)^{1/q})$ is an abelian Galois extension of $\mathbb{Q}_p$, also leading to the desired extension of degree $q$. That leaves the case of $q$ not dividing $p-1$ but still dividing $p^f-1$; in these cases, the extension is not Galois. Can we say anything about intermediate extensions of degree $q$ in this case?