I know that if we have $L/K$ and $K'/K$ two extensions inside the algebraic closure $\bar K/K$, and $L'=LK'$. Then we have:
$$L/K \:\text{tamely ramified} \Rightarrow L'/K' \:\text{tamely ramified}.$$
Why does it follow that a subextension of a tamely ramified extension is tamely ramified? I see that if we have a tower of extensions $L/K'/K$ then $L/K'$ is tamely ramified (because using the theorem above, $L'=LK'=L$).
Let $M/K$ be a subextension of the tamely ramified extension $L/K$. To prove that $M/K$ is tamely ramified we must show that $(e(M/K),p)=1$, where $p=\operatorname{char}(k)$ where $k$ is the residue field of $K$. By assumption we have that $(e(L/K),p)=1$.
Recall the tower formula for ramification indices: $$e(L/K)=e(L/M)e(M/K).$$ So, if $(e(M/K),p)>1$ then $p\mid e(M/K)$ because $p$ is prime, which implies that $p|e(L/K)$, which is a contradiction.
Therefore $(e(M/K),p)=1$ and $M/K$ is tamely ramified.