Let $\mathbb{Z}$ be defined with the usual metric. Show that every subset of $\mathbb{Z}$ is open in $\mathbb{Z}$.
I can write $\mathbb{Z}$ as $\bigcup_{n=0}^{\infty} \{n\}$ and then letting $B(n, \frac{1}{2})$ I get that $B(n, \frac12) \cap \mathbb{Z} = \{n\} $. So it seems that I can write any element in $\mathbb{Z}$ as the intersection of two open sets and thus it would be open? How can I formalize this, I seem to have a bit hard time doing that...?
By the usual metric on $\Bbb Z$ you seem to mean a metric (induced by the Euclidean distance function?) which would induce the discrete topology. Once you get points to be open, it's discrete. But then every single subset of $\Bbb Z$ would be open. So you are done.
When you write $B(n, \frac 1 2) \cap \Bbb Z$, you seem to have in mind the idea of a "ball" in $\Bbb R$.
So you should forget the fact that $\Bbb Z$ is a subspace of $\Bbb R$. It is just a set, together with a metric function $d: \Bbb Z \times \Bbb Z \rightarrow \Bbb R_{\geq 0}$, sending $(x, y)$ to $|x - y|$ (viewed as a non-negative real number).
By definition, a subset $X$ of $\Bbb Z$ is open if for every element $x \in X$, there exists a positive real number $r$ such that the set $B(x, r) = \{y \in \Bbb Z: d(x, y) < r\}$ is contained in $X$.
You can then show that every subset $X$ is open by choosing $r$ above to be $\frac 1 2$.
