Let $p$ be a prime number and $\mathbb{Q}_p$ be the $p-$adic rational field. Let $E/\mathbb{Q}_p$ be a fixed finite extension. On this site, I define a finite extension $F/E$ is "good" if there exists an intermediate extension $K/E$ of $F/E$ such that $K/E$ is totally ramified and $F/K$ is unramified.

My question is, for any finite extension $L/E$, does there exist a "good" extension $F/E$, such that $L$ is contained in $F$? Since we only care about the inclusion $L\subset F$, we can enlarge $L$, such that $L/E$ is a Galois extension.

If $L/E$ is an abelian extension, the answer to my question is yes. From class field theory, we can construct the Lubin-Tate extension $E_\pi/E$, which is totally ramified. The maximal abelian extension of $E$ is obtained by the composite of $E_\pi$ and the maximal unramified extension of $E$. Hence we can find a "good" extension $F/E$ such that $F\supset L$.

Since this question looks rather fundamental, I wonder if this question was considered and solved by anyone else before. Any help will be appreciated.

As commented by @re'em waxman, I'd like to make it more precise. Let $\mathbb{Q}_p^{\mathrm{al}}$ denote the algebraic closure. The comment by @re'em waxman suggested me to find an algebraic extension $K/E$, such that $K\cap E^{\mathrm{ur}}=E$(it is the same as $K/E$ is totally ramified) and $KE^{\mathrm{ur}}=\mathbb{Q}_p^{\mathrm{al}}$. From infinite Galois theory, it is equivalent to finding a closed subgroup $H$of the absolute Galois group $G_E$, such that $I_E\cap H=\{1\} $ and $I_E H=G_E$, where $I_E$ denotes the inertia group of $E$. In other words, it is equivalent to finding a subgroup $H$, such that the composite of natural maps $H\hookrightarrow G_E\twoheadrightarrow G_{E}/I_E$ is an isomorphism of profinite groups. Finally, it is equivalent to the following short exact sequence of profinite groups splits: $$0\longrightarrow I_E\longrightarrow G_E\longrightarrow G_E/I_E\cong \hat{\mathbb{Z}}\longrightarrow 0.$$ One suggested $H$ is taken to be the closure of the subgroup generated by any chosen lifted Frobenius $\tilde{\varphi}$. But I don't know if this is a correct $H$. In other words, does the group homomorphism $\mathbb{Z}\to G_E$ by $1\mapsto \tilde{\varphi}$ extend to a continuous group homomorphism $\hat{\mathbb{Z}}\to G_E$?

Since $G_E$ is profinite, $$ \mathrm{Hom}(\hat{\mathbb{Z}},G_E)= \mathrm{lim}_{\leftarrow}\mathrm{Hom}(\hat{\mathbb{Z}},G_E/U)\cong \mathrm{lim}_{\leftarrow}G_E/U=G_E,$$ where the inverse limit runs through open normal subgroups $U$ of $G_E$. We see any homomorphism $\mathbb{Z}\to G_E$ extends continuously to $\hat{\mathbb{Z}}$, as pointed out by @David Loeffler. So my problem is solved. Thank you for your help again.