I want to check $p^n \Bbb{Z}_p$ is closed subgroup of $ \Bbb{Z}_p$. It is clear that $p^n \Bbb{Z}_p$ is subgroup of $ \Bbb{Z}_p$, I want to prove $p^n \Bbb{Z}_p$ is closed in $\Bbb{Z}_p$.
So, I only need to check $\Bbb{Z}_p\setminus p^n \Bbb{Z}_p$ is open in $p^n \Bbb{Z}_p$. Now,basis of $ \Bbb{Z}_p$ is $a +p^e \Bbb{Z}_p$($e≧0$) from the definition of p-adic topology. So, I only need to check $\Bbb{Z}_p\setminus p^n \Bbb{Z}_p$ can be rewritten as union of $a +p^e \Bbb{Z}_p$($e≧0$).
How can I write $\Bbb{Z}_p\setminus p^n \Bbb{Z}_p$ as union of $a +p^e \Bbb{Z}_p$($e≧0$)?
Let $\alpha \in \mathbb{Z}_p \setminus p^n\mathbb{Z}_p$. Let $U = \alpha + p^n\mathbb{Z}_p$. Then $U$ is basic open by your definition, and we have $\alpha \in U\subseteq \mathbb{Z}_p\setminus p^n\mathbb{Z}_p$. To see the last inclusion, suppose that $x \in U \cap p^n\mathbb{Z}_p$. Since $x \in U$ we have $x \equiv \alpha \pmod{p^n}$, and since $x \in p^n\mathbb{Z}_p$ we have $x \equiv 0 \pmod{p^n}$. Therefore, we get $\alpha \equiv 0 \pmod{p^n}$, so $\alpha \in p^n\mathbb{Z}_p$, which contradicts the definition of $\alpha$.
I think the point of Captain Lama's comment was that $p^n\mathbb{Z}_p$ is a closed ball (or disc/disk, depending on your terminology) by definition, so it is closed by standard theory.