I'm trying to understand a proof of the following:
Let $L$ and $M$ be Abelian extensions of $K$ (a number field). Then $L\subset M$ if and only if there is a modulus $\mathfrak{m}$, divisible by all primes of $K$ ramified in either $L$ or $M$, such that
$$P_{K,1}(\mathfrak{m}) \subset \ker(\Phi_{M/K,\mathfrak{m}}) \subset \ker(\Phi_{L/K,\mathfrak{m}}).$$
(Here $P_{K,1}(\mathfrak{m}) = \{\alpha\in K^*\mid \alpha\equiv^*1\mod{m}\}$, and $\Phi$ is the Artin map.)
Note: this is the same question as this one, $\color{red}{\text{but I'm asking for clarification of a particular proof that does not involve ideles }}$ $\color{red}{\text{(as the proof in the linked question does).}}$
The proof I'm looking at goes as follows: First assume $L\subset M$, and let $r:\text{Gal}(M/K)\to \text{Gal}(L/K)$ be the restriction map. Choose a modulus $\mathfrak{m}$ for which both $\ker(\Phi_{L/K,\mathfrak{m}})$ and $\ker(\Phi_{M/K,\mathfrak{m}})$ are congruence subgroups modulo $\mathfrak{m}$. Since $r\circ \Phi_{M/K,\mathfrak{m}} = \Phi_{L/K,\mathfrak{m}}$, it follows that $$\ker(\Phi_{M/K,\mathfrak{m}}) \subset \ker(\Phi_{L/K,\mathfrak{m}}).$$ That part is clear.
For the other direction, suppose that the inclusion condition in the statement holds, and let $\ker(\Phi_{L/K,\mathfrak{m}})\subset I_K(\mathfrak{m})$ correspond to $H\subset \text{Gal}(M/K)$ under the Artin map $\Phi_{M/K,\mathfrak{m}}$. Then $H$ corresponds to some intermediate field $K\subset\tilde{L}\subset M$. $\color{red}{\text{The first part of the proof, applied to }}$ $\color{red}{\tilde{L}\subset M}\color{red}{\text{, shows that }}$ $\color{red}{\ker(\Phi_{\tilde{L}/K,\mathfrak{m}})=\ker(\Phi_{L/K,\mathfrak{m}}).}$ Then uniqueness proves that $L = \tilde{L}$, and we are done.
The sentence in red is the one I'm confused about. I just don't see how the first part of the proof is relevant. One can use it to show that $\ker(\Phi_{M/K,\mathfrak{m}}) \subset \ker(\Phi_{\tilde{L}/K,\mathfrak{m}})$, but then what? Why does that imply that the kernels are equal?