Is the set $\{2,3,5\}$ connected?

I read somewhere that a connected set in the digital line topology is a subset of consecutive integers. Recall that the digital line topology is a topology on $$\mathbb{Z}$$ with basis elements $$\{n\}$$ for $$n$$ odd and $$\{n-1, n, n+1\}$$ for $$n$$ even. However, if we consider the set $$\{2,3,5\}$$, then this set must be disconnected based on the above statement, as it is missing the integer $$4$$. In order to show that this set is disconnected, we must show that it has a separation (or that it is the union of two disjoint nonempty open sets). $$\{5\}$$ is open in the digital line topology, as $$5$$ is odd. However, for the case of the set $$\{2,3\}$$ it's trickier. We can show that this set is not open, as we cannot construct an open neighborhood around $$\{2\}$$ such that the neighborhood is entirely contained in the two-tuple set. How do I show that this set is indeed disconnected?

The set $$\{2,3\}$$ is open, since it is equal to $$\{1,2,3\}\cap\{2,3,5\}$$ and $$\{1,2,3\}$$ is open set in $$\mathbb Z$$.
To say that a subset is disconnected is to say that it is disconnected under the subspace topology. To show that $$\{2,3,5\}$$ is disconnected, you can show that it is a disjoint union of sets that are open under the subspace topology. Remember that a set $$S$$ is open under the subspace topology here if $$S=T\cap\{2,3,5\}$$ where $$T$$ is open in the original topology, in this case, the digital line topology. Then $$\{2,3,5\}=\{2,3\}\cup \{5\}$$, the former of which is open because $$\{2,3\}=\{1,2,3\}\cap\{2,3,5\}$$, and $$\{1,2,3\}$$ is open in the digital line topology, and the latter is open as you pointed out.