[proofplan]
We show $K^{\mathrm{ab}} = K^{\mathrm{ur}} L$ by proving that every finite abelian extension of $K$ is contained in $K^{\mathrm{ur}} L$. The key tool is the norm group characterization from local class field theory: a finite abelian extension $M/K$ corresponds to the open subgroup $N(M/K) \subseteq K^\times$, and $M \subseteq K^{\mathrm{ur}} L$ if and only if $N(K^{\mathrm{ur}} L_n/K) \subseteq N(M/K)$ for some $n$. We identify $N(K^{\mathrm{ur}} L_n/K)$ as the group generated by $\pi_K^{f_n}$ and $U_K^{(n)}$ (where $f_n$ depends on the unramified part), and show that every open subgroup of finite index in $K^\times$ contains such a group.
[/proofplan]
[step:Show every finite abelian extension is contained in $K^{\mathrm{ur}} L_n$ for some $n$]
Let $M/K$ be any finite abelian extension. By local class field theory, $M$ corresponds to the open subgroup $N(M/K)$ of finite index in $K^\times$. Write $K^\times = \langle \pi_K \rangle \times U_K$. Since $N(M/K)$ has finite index, it contains $\pi_K^m$ for some $m \geq 1$ (the image of $N(M/K)$ under $v_K: K^\times \to \mathbb{Z}$ is a subgroup of finite index in $\mathbb{Z}$, hence $m\mathbb{Z}$ for some $m$). Also, $N(M/K) \cap U_K$ is an open subgroup of $U_K$, so it contains $U_K^{(n)}$ for some $n \geq 1$ (since the $U_K^{(n)}$ form a fundamental system of open neighbourhoods of $1$ in $U_K$).
Therefore $\langle \pi_K^m \rangle \times U_K^{(n)} \subseteq N(M/K)$.
[guided]
Why is $N(M/K)$ open? Local class field theory establishes a bijection between finite abelian extensions $M/K$ and open subgroups of finite index in $K^\times$ (via the norm map). The openness comes from the topology on $K^\times$: the norm map $N_{M/K}: M^\times \to K^\times$ is continuous, and $N(M/K) = N_{M/K}(M^\times)$ is open because it contains an open subgroup. Concretely, $N(M/K) \supseteq N_{M/K}(U_M^{(k)})$ for large $k$, and norm maps on unit groups preserve openness.
The key point is that the unit filtration $U_K \supset U_K^{(1)} \supset U_K^{(2)} \supset \cdots$ gives a neighbourhood basis at $1$, and any open subgroup of $U_K$ must contain some $U_K^{(n)}$.
[/guided]
[/step]
[step:Identify the norm group of $K^{\mathrm{ur}} L_n$]
The extension $K^{\mathrm{ur}}/K$ is the maximal unramified extension, and $N(K^{\mathrm{ur}}_m/K) = \langle \pi_K^m \rangle \times U_K$ for the unramified extension of degree $m$ (since $\operatorname{Frob}_K$ has order $m$ in $\operatorname{Gal}(K^{\mathrm{ur}}_m/K)$, and the norm map on units is surjective for unramified extensions). The extension $L_n/K$ is totally ramified with $N(L_n/K) = \langle \pi_K \rangle \times U_K^{(n)}$ by the [Norm Group of Lubin--Tate Extensions](/theorems/2395).
For the compositum $K^{\mathrm{ur}}_m \cdot L_n$, the norm group is
\begin{align*}
N(K^{\mathrm{ur}}_m \cdot L_n / K) = N(K^{\mathrm{ur}}_m/K) \cap N(L_n/K) = (\langle \pi_K^m \rangle \times U_K) \cap (\langle \pi_K \rangle \times U_K^{(n)}) = \langle \pi_K^m \rangle \times U_K^{(n)},
\end{align*}
where the first equality uses the [Containment of Norm Groups](/theorems/2389) (the norm group of a compositum is the intersection of the norm groups).
[/step]
[step:Conclude $M \subseteq K^{\mathrm{ur}} L_n$ and hence $K^{\mathrm{ab}} = K^{\mathrm{ur}} L$]
From the first step, $\langle \pi_K^m \rangle \times U_K^{(n)} \subseteq N(M/K)$, and from the second step, $N(K^{\mathrm{ur}}_m \cdot L_n / K) = \langle \pi_K^m \rangle \times U_K^{(n)}$. Therefore
\begin{align*}
N(K^{\mathrm{ur}}_m \cdot L_n / K) \subseteq N(M/K).
\end{align*}
By the [Containment of Norm Groups](/theorems/2389), this implies $M \subseteq K^{\mathrm{ur}}_m \cdot L_n \subseteq K^{\mathrm{ur}} \cdot L$.
Since every finite abelian extension $M/K$ is contained in $K^{\mathrm{ur}} L$, and $K^{\mathrm{ab}}$ is the union of all finite abelian extensions:
\begin{align*}
K^{\mathrm{ab}} = \bigcup_{M/K \text{ finite abelian}} M \subseteq K^{\mathrm{ur}} L.
\end{align*}
The reverse containment $K^{\mathrm{ur}} L \subseteq K^{\mathrm{ab}}$ holds because $K^{\mathrm{ur}}/K$ is abelian (its Galois group is $\hat{\mathbb{Z}}$, which is abelian) and each $L_n/K$ is abelian (by the [Galois Group of Lubin--Tate Extensions](/theorems/2394)), so $K^{\mathrm{ur}} L = K^{\mathrm{ur}} \cdot \bigcup_n L_n$ is a union of compositums of abelian extensions, hence abelian over $K$.
Therefore $K^{\mathrm{ab}} = K^{\mathrm{ur}} L$.
[/step]
