Let $p$ be a prime, let $\pi_n : \mathbb{Z}/p^{n+1}\mathbb{Z} \to \mathbb{Z}/p^n\mathbb{Z}$ be the obvious projection and $$\mathbb{Z}_p = \{\, x \in \Pi_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z} \,|\, \forall k\geq 1: \pi_k(x_{k+1}) = x_k \,\}$$ the p-adic integers.
Endow $\mathbb{Z}/p^n\mathbb{Z}$ with the discrete topology and $\mathbb{Z}_p$ with the product topology.
Now, if I'm not mistaken, open sets in $\mathbb{Z}_p$ are now given as (unions of) $\Pi_n S_n$ with only finitely many $S_n$ not equal to $\mathbb{Z}/p^n\mathbb{Z}$. However, this is clearly inconsistent with the requirement on $x\in\mathbb{Z}_p$ that $\pi_n(x_{n+1}) = x_n$ (since we can pick $n$ large enough so we can choose, say, $\pi_n(x_{n+1}) = x_n+1$, since only finitely many $S_n$ place any effective restrictions on the choice of $x_n$ and $x_{n+1}$).
So what do open sets in the product topology on $\mathbb{Z}_p$ look like?
As mentioned in a comment, there is an important subtlety: $\Bbb Z_p$ is just a subset of the direct product, so "the product topology" on it makes no sense. Instead, one endows it with the subspace topology of the product topology.
Spelled out, this means that a set $X \subseteq \Bbb Z_p$ is open iff it can be written as $\Bbb Z_p \cap \tilde X$ where $\tilde X$ is an open set of $\prod_{n\ge 1} \Bbb Z/p^n$ with the product topology. Spelled out further and using your correct description of the product topology, that means that $X \subseteq \Bbb Z_p$ is open iff for every $x \in X$ there is some set of the form $\Bbb Z_p \cap \prod_n S_n$ with finitely many $S_n$ not equal to $\Bbb Z/p^n$ containing $x$ and contained in $X$. Now say $k_{x, X}$ is a number such that all those $S_n = \Bbb Z/p^n$ for $n \ge k$, you see this means that with $x$ also $x + p^{k_{x, X}} \Bbb Z_p \subseteq X$, and that is one standard description of open sets in $\Bbb Z_p$.
Loosely speaking, a basic open set containing a consistent sequence $x =(x_k)_k \in \Bbb Z_p$ is of the form "every consistent sequence $(y_k)_k$ such that $y_k=x_k$ for (at least) finitely many $k$".