Is any subset of $\mathbb{Z}$ open?

Question

My question is to consider $\mathbb{Z} \subseteq \mathbb{R}$, with the subspace metric. What are the open subsets of $\mathbb{Z}$? The answer to this is that any subset of $\mathbb{Z}$ is open. I am struggling to prove this. Can anyone help me out? I want to prove this using the metric space characterization of open sets:

Let $(X,d)$ be a metric space. A subset $U \subseteq X$ is an open set if for each $x \in U$ there exists $r > 0$ such that $B(x,r) \subseteq U$. Where $B(x,r)$ is the open ball with radius $r$ and centre $x$.

Answer 1

Hint:

Since every set can be written as a union of singletons, you should try to prove that singletons are open sets.

Hint 2:

For a singleton $\{n\}\subseteq\mathbb Z$, try to find an open set $O$ in $\mathbb R$ such that $O\cap\mathbb Z=\{n\}$

Hint 3:

Think of simple candidates for $O$.

Answer 2

Hint: remember that the open ball $B(x, r)$ of radius $r$ about $x$ depends on the subspace: for example, in the subspace $\Bbb{Z}$ of the metric space $\Bbb{R}$, $B(0, 2)$ means something different from what it means in $\Bbb{R}$: in $\Bbb{Z}$ it means all the integers at distance less than $2$ from $0$, not all the real numbers with that property. So in $\Bbb{Z}$, $B(0, 2)$ has just three elements. Now can you think of a $d > 0$ such that $B(i, d)$ (in $\Bbb{Z}$) is contained in any subset that contains the integer $i$?