Let $K\subset \mathbb{C}$ be a subfield, and $q$ a prime number. Suppose, for all field extension $L/K$ with finite degree, $q|[L:K] $. Then, for all finite extension $L$, there exists $r\in \mathbb{N} \cup \{0\}$ s.t., $[L:K]=q^r$.
If there exists $L$ with $[L:K]=nq^r,\;(q\not \mid n)$, its Galois closure is finite extension. So, I can prove by contradiction when Galois group is abelian, since structure theorem of finite abelian group. But I can't prove general case.
If there exists $L$ with $[L:K] = nq^r$, then its Galois closure $L'$ is a finite extension and $[L':K] = n'q^{r'}$ for some $n'$ and $r'$ with $q \nmid n'$ (and of course, $n| n'$ and $r' \geq 0$). The Galois group $Gal(L' : K)$ is of order $n'q^{r'}$.
By Sylow's theorems, $Gal(L':K)$ has a subgroup of order $q^{r'}$. This subgroup corresponds to an intermediate field $E$ (with $L' \supset E \supset K$) such that $[L' : E] = q^{r'}$. But then, $[E : K] = n'$, contradicting the assumption that $q | [E : K]$.