I am given the following definition (without the proof or technical details).
and I need to understand that
I tried the following:
Since $P_{K,1}(\mathfrak{m}) \subseteq \ker(\Phi_{\mathfrak{m}})$ we have: $$\begin{array}{ccccc} \frac{I_K(\mathfrak{m})}{\ker(\Phi_{\mathfrak{m}})} & \subseteq & \frac{I_K(\mathfrak{m})}{P_{K,1}(\mathfrak{m})} & & \\ & & & & \\ \cong & & \cong & & \\ & & & & \\ Gal(L/K) & \subseteq & Gal(K(\mathfrak{m})/K) & \underbrace{\subseteq}_{Lemma: \ true?} & Gal(K(\mathfrak{n})/K) \\ & & & & \\ \cong & & & & \cong \\ & & & & \\ \frac{I_K(\mathfrak{n})}{\ker(\Phi_{\mathfrak{n}})} & & \subseteq & & \frac{I_K(\mathfrak{n})}{P_{K,1}(\mathfrak{n})} \\ \end{array}$$
I am tempted to conclude from this that $P_{K,1}(\mathfrak{n}) \subseteq \ker(\Phi_{\mathfrak{n}})$, but this is an invalid deduction. Consider the example:
$$\frac{\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2}{0 \times \mathbb{Z}_2 \times \mathbb{Z}_2} \subseteq \frac{\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2}{\mathbb{Z}_2 \times 0 \times 0}$$
Nonetheless, I feel that my train of thought was not completely wrong and that under a certain extra assumption the inclusion must hold...
Your proof does not work (for the reason John has explained in the comment). Note that the statment you want to prove does not involve Artin's reciprocity theorem itself but let's see how it fits together: Part (ii) of the theorem tells us, that whenever we make the modulus $\mathfrak m$ large enough the Artin map $$ \Phi_\mathfrak m \colon I_K(\mathfrak m) \to \operatorname{Gal}(L|K)$$ satisfies $P_{K,1}(\mathfrak m) \subseteq \ker(\Phi_\mathfrak m)$. We now want to show that once this holds for $\mathfrak m$, it also holds for every modulus $\mathfrak n$ with $\mathfrak m \mid \mathfrak n$.
(Such a modulus is usually called an admissible modulus. With this notion Artin's theorem tells us that (among other things) there exists an admissible modulus for every abelian extension. What you want to prove is: If there exists an admissible modulus, then there are infinitely many. Note that Artin does not give a criterion for a modulus to be admissible!)
Now let's prove this. Let $(\alpha) \in P_{K,1}(\mathfrak n)$ coprime to $\mathfrak n$ such that $\alpha \equiv 1 \bmod \mathfrak n_0$ and $\sigma(\alpha) > 0$ for every real infinite prime dividing $\mathfrak n_\infty$ (this is just the definition of $P_{K,1}$). Now as $\mathfrak m_0$ divides $\mathfrak n_0$, $\mathfrak m_\infty$ divides $\mathfrak n_\infty$ and $(\alpha)$ is coprime to $\mathfrak m_0$ we have $(\alpha) \in P_{K,1}(\mathfrak m)$. Thus $(\alpha) \in \ker(\Phi_\mathfrak m)$. Now what is the connection between $\Phi_\mathfrak n$ and $\Phi_\mathfrak m$? We have $I_K(\mathfrak n) \subseteq I_K(\mathfrak m)$ and $\Phi_\mathfrak n$ is just the restriction of $\Phi_\mathfrak m$ to $I_K(\mathfrak n)$. Since $(\alpha) \in I_K(\mathfrak n)$ this implies $\Phi_\mathfrak n((\alpha)) = \Phi_\mathfrak m((\alpha)) = \mathrm{id}_K$ and therefore $(\alpha) \in \ker(\Phi_\mathfrak n)$. This finishes the proof. Note that we have just proven $$ P_{K,1}(\mathfrak n) \subseteq P_{K,1}(\mathfrak m) \subseteq \ker(\Phi_\mathfrak m) \subseteq \ker(\Phi_\mathfrak n). $$