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My proof is:

Let $x$ be any point in $\mathbb{N}$, and let $\beta$ be the closest number (such as a factor or irrational number) to $x$, such that $\beta \notin \mathbb{N}$ and $\beta > x$.

Let $B_\epsilon(x)$ be the epsilon ball around $x$ in $\mathbb{N}$. Let $\epsilon = \beta - x$.

Clearly $\forall x \in \mathbb{N}: \exists \epsilon >0, \epsilon \in \mathbb{R}: B_\epsilon \left({x}\right) \subseteq \mathbb{N}$. So we have that all singleton sets in $\mathbb{N}$ are open.

All subsets of $\mathbb{N}$, except the empty set, consist of a union of many such $x \in \mathbb{N}$ and since all unions of open sets are open, we have that all subsets of $\mathbb{N}$ are open.

This means that the complement to every subset in $\mathbb{N}$ is closed. However, since the complement to any subset of $\mathbb{N}$ is itself a subset of the space, we have that all subsets of $\mathbb{N}$ are closed.

Therefore, all subsets of $\mathbb{N}$ are clopen.

Is this correct?

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    Surely $x + \frac{\beta - x}{2}$ is closer to $x$ than $\beta$ is, showing that no such $\beta$ exists. If $x$ is $3$ what is $\beta$ supposed to be? – 2017-02-07
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    I don't really understand your definition of $\beta$. Why not just take $\varepsilon=\frac{1}{2}$? – 2017-02-07
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    I see this now, very dumb mistake. Thanks. – 2017-02-07
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    @carmichael561 if you'd like to post this as an answer, I will gladly accept it. – 2017-02-07

2 Answers 2

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There is no closest number to $x$, there is always another number between $x$ and any $\beta$ you come up with.

However you have the right idea about showing the singletons are open, just use $\epsilon = \frac{1}{2}$.

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Just show all singletons are open, which is easy, as all natural numbers are at least 1 apart: $\{n\} = B_\frac{1}{2}(n)$ where the ball is taken in the restricted metric. If all singletons are open, all subsets $A \subseteq \mathbb{N}$ are open, as $A = \cup\{\{n\}: n \in A\}$ is a union of open sets, so open.

This also makes all subsets closed, as all complements are open as well.