Let $K$ be a local field and $\pi$ a uniformizer of $O_K$. In both Milne's notes and these, it is proven that the local Artin map must be unique if it exists, assuming the local existence theorem. For reference, these are the statements:
Local Artin map: $\phi: K^\times\rightarrow \operatorname{Gal}(K^{ab}/K)$ that restricts to an isomorphism $\phi_L: K^\times/N(L^\times)\rightarrow\operatorname{Gal}(L/K)$ for each finite abelian $L/K$, AND if $L/K$ is unramified, the image of $\pi$ acts by the Frobenius of $\operatorname{Gal}(L/K)$.
Local existence: Every finite index open subgroup of $K^\times$ is the norm group of some finite abelian extension of $K$.
The proofs given in both can be briefly summarized like this: We construct $K_{\pi, n}$ to be the fields corresponding to the subgroups $(1+\mathfrak{p}^n)\langle \pi\rangle$ and define $K_\pi$ to be their union. Then it is claimed that $K^{ab}=K_\pi K^{unr}$. It is then shown through the properties of the local Artin map that the action of $\pi$ on both these subfields is determined - trivial on $K_\pi$ and Frobenius on $K^{unr}$, and the result follows since $K^\times$ is generated by its uniformizers.
My confusion arises from the claim that $K^{ab}=K_\pi K^{unr}$. This fact is indeed proven earlier, but a) with a different definition of $K_\pi$ and b) SEEMING to assuming the full strength of local CFT! (In particular, it follows from the isomorphism of the profinite completions of $K^\times$ and $\operatorname{Gal}(L/K)$; more on this later.) Indeed, in the previous definition of $K_\pi$, it is essentially defined as the fixed field of $\pi$, from which the key statement that the action of $\pi$ is trivial on $K_\pi$, is tautological.
This motivates me to try to show that if we define $K_\pi$ as the union of all finite totally ramified extensions of $K$ with $\pi\in N(L^\times)$, then $K^{ab}=K_\pi K^{unr}$. If this is true, then we know the action of $\pi$ is fixed on each piece. However, this statement is not clear. For instance, if we define $M$ to be the union of ALL finite totally ramified extensions of $K$, then we do have $K^{ab}=MK^{unr}$, but in general I don't think $M=K_\pi$.
After reading the proofs carefully again, it appears to me that the proof that $K^{ab}=K_\pi K^{unr}$ does not need the full strength of local CFT (namely, does not need uniqueness). I believe that the existence of the Artin map gives the isomorphism $\widehat{K^\times}\cong \operatorname{Gal}(K^{ab}/K)$; uniqueness is used merely to show this isomorphism is canonical. Then we can split $\widehat{K^\times}$ up and use the Galois correspondence to reach the desired statement.
Now this seems to prove uniqueness just fine - but this still doesn't clear up my confusion, because I don't see the point of defining these $K_{\pi, n}$s! I gather that they are used for Lubin-Tate theory, but they seem to be completely irrelevant to the proof.
To summarize, my two questions are: 1) Is my outline of a proof valid, and 2) Why did they include the $K_{\pi, n}$s in this proof/am I interpreting their proof correctly?
Edit: I don't know why I originally included the now-struck-through paragraph; it was part of a flawed earlier attempt which has nothing to do with my actual question (which remains the same).