Why is this function an embedding? We have the canonical function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p, x \mapsto (\overline{x})_{k \in \mathbb{N}_0}=(\overline{x}, \overline{x}, \overline{x}, \dots )$. 
The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding. We interpret with it $\mathbb{Z}$ as a subset of $\mathbb{Z}_p$. 
Proof: 
Let $x \in \mathbb{Z}$ with $\epsilon_p(x)=(\overline{x})_k=0$, i.e. $x \equiv 0 \mod{p^{k+1}}$ for all $k \in \mathbb{N}_0$. The only integer number that is divisible by any high $p$-power is zero, so it holds $x=0$. Thus, $\epsilon_p$ is an embedding. 
Could you explain me the above proof?
 A: Let's look at the definition of $\mathbb{Z}_p:$
$$
\mathbb{Z}_p := \{(x_n)_{n \geq 0} \in \Pi_{n \in \mathbb{N}_0}\mathbb{Z}/p^{n+1}\mathbb{Z} | x_n = \eta_{mn}(x_m), \forall m \geq n  \}
$$
where for $m \geq n, \eta_{mn}: \mathbb{Z}/p^{m}\mathbb{Z} \rightarrow \mathbb{Z}/p^{n}\mathbb{Z}$ is the natural map induced by the inclusion $p^m\mathbb{Z} \subseteq p^n \mathbb{Z}.$
Now consider the map $\epsilon_p : \mathbb{Z} \rightarrow \mathbb{Z}_p$ defined by $x \mapsto (x_k)_{k \geq 0}$ where $x_k$ is the image of $x$ in $\mathbb{Z}/p^{k}\mathbb{Z}.$ Then $\epsilon_p$ is an abelian group homomorphism. Suppose that for some $x, y \in \mathbb{Z}, \epsilon_p(x) = \epsilon_p(y).$ Then we have $x_k = y_k$ in $\mathbb{Z}/p^{k +1}\mathbb{Z}, \forall k \geq 0,$ i.e. $x - y$ is divisible by $p^{k+1}, \forall k \geq 0.$ Thus $x = y$ and consequently $\epsilon_p$ is an embedding.
A: In general, to show a ring homomorphism $\varphi$ is injective, it is equivalent to show that $\varphi(0) = 0$, since $$\varphi(x) = \varphi(y) \iff \varphi(x-y)=0$$
so $\varphi$is injective if and only if $$x=y \iff \varphi(0) = 0$$
The proof has skipped the step of showing that your function $\epsilon_p$ is a ring homomorphism, and is just using the above fact to show it is injective.

We need to show that $\epsilon_p(x) = 0 \implies x = 0$. But by definition, $$\epsilon_p(x) = 0 \iff (\overline x, \overline x,\overline x,\ldots)= (0,0,0\ldots) \iff x \equiv 0 \pmod {p^{k}} \ \forall k \in \mathbb N$$ So $p^k \mid x$ for every $k$, and this can only occur if $x =0$.
