Let $S$ be a monoid. A non-empty set $A$ is called a right $S$-act if there exists a map $A\times S\to A$ such that $(as)t=a(st)$ and $a1=a$ for all $s,t\in S$ and $a\in A$.
A sub-act of a right $S$-act $A$ is a non-empty subset of $A$ which is closed under the action of $S$.
A right S-act is called simple if it has no trivial subact.
For a prime $p$, $\mathbb{Z}_p$ is a right $\mathbb{N}$-act when we define the map $\mathbb{Z}_p \times \mathbb{N}\to \mathbb{Z}_p$ by $(\bar{x},n)\mapsto n\bar{x}$. My question is that: is $\mathbb{Z}_p$ a simple $\mathbb{N}$-act?
What I've tried: Let $A\neq \{ 0\}$ be a sub-act of $\mathbb{Z}_p$. Then there exists $\bar{0}\neq \bar{x}\in A$. Now if there exists $n_0\in \mathbb{N}$ such that $n_0 x\overset{p}{\equiv}1$, then $\bar{1}=n_0 \bar{x}\in A$. Then for all $n\in \mathbb{N}$, $\bar{n}=n\bar{1}\in A$ hence $A=\mathbb{Z}_p$.
Is there $n_0\in \mathbb{N}$ such that $n_0 x\overset{p}{\equiv}1$?
As you noticed correctly, it is enough to show that for all $x\not\equiv 0 \bmod p$ there exists an $n_0\in \mathbb{N}$ such that $n_0 x \equiv 1 \bmod p$.
This follows directly from $\gcd(x, p) = 1$ whenever $x \not\equiv 0 \bmod p$.
For more information see https://en.wikipedia.org/wiki/Modular_multiplicative_inverse or, for example, If $q$ is coprime to $a$ then $a\mid nq-1,\,$ so $q$ is invertible mod $a$ .