If you want to show that all subsets are open, the easiest way to do so is to show that singletons are open (and hence all subsets would be - why?) : So just take a singleton $\{n\} \subset \mathbb{N}$, then for $r = 1/2$, then open ball
$$
B(n,r)\cap \mathbb{N} = \{n\}
$$
Hence, $\{n\}$ is open, and you're done!
Edit: Your argument is almost correct. You choose a point $x\in \mathbb{N}$ and you want to check where $d(x,S) = d(x,S^c) = 0$. There are two cases, either $x\in S$, or $x\in S^c$. In the first case,
$$
d(x,S) = 0, \text{ but } d(x,S^c) \geq 1
$$
as you have observed. And in the second case, the reverse is true.
Your statement that $d(x,S) = d(x,S^c) = 1$ is not true for any $x\in \mathbb{N}$!
The following books explicitly take the position that $\operatorname{diam}\varnothing =0$:
- C. Kuratowski, Topology, vol.I
- M. H. A. Newman, Elements of the topology of plane sets of points
The following books explicitly take the position that $\operatorname{diam}\varnothing =-\infty$:
- G. F. Simmons, Introduction to topology and modern analysis
- M. Ó. Searcóid, Metric spaces
(I never heard of either of these before Google Books search brought them up.)
The following books explicitly restrict the definition of diameter to nonempty sets:
- W. Rudin, Principles of mathematical analysis
- H. L. Royden, Real analysis.
- K. Falconer, Fractal geometry
It seems that W. Sierpiński, General topology, belongs to the second or third category, because the author says on page 110: "Thus the diameter of every non-empty set contained in a metric space is a uniquely defined real non-negative number, finite or infinite". But it's not very clear what Sierpiński's intention was when writing this.
Many books do nothing of the above: they define the diameter of a set as supremum of pairwise distances, and offer no further details.
If you allow the diameter of the empty set to be $−\infty$, does it lead to problems?
The definition of Hausdorff measure would become awkward. For example, the 1-dimensional measure involves the infimum of $\sum \operatorname{diam} U_i$ over certain families of sets. If $\operatorname{diam}\varnothing =-\infty$, we'd be able to make the infimum $-\infty$ by throwing in the empty set. (Note that the Wikipedia article explicitly says that $\operatorname{diam}\varnothing =0$). One can try to fix this by requiring $U_i$ to be nonempty, but then the measure of empty space becomes a special case (and the measure of $\varnothing$ definitely needs to be $0$).
Another issue is the inequality
$$
\operatorname{diam}(A\cup B)\le \operatorname{diam}A+\operatorname{diam}B+\operatorname{dist}(A,B)
$$
which should hold for all $A,B$. Suppose $B$ is empty but $A$ is not. The right-hand side becomes undefined due to presence of
$\operatorname{diam}\varnothing =-\infty$ and $\operatorname{dist}(A,\varnothing)=+\infty$. (And the latter definitely needs to be $+\infty$.)
Third issue: if one applies a metric transform, i.e., replaces metric $d$ with $\varphi(d)$ where $\varphi $ is an increasing concave function, the diameters of sets should transform accordingly. With $-\infty$ in the mix, one is led to awkward conventions ($\sqrt{-\infty}=-\infty$?).
That said, I can imagine some arguments in favor of $\operatorname{diam}\varnothing =-\infty$. One is that the following statement becomes true:
In a complete metric space, each decreasing sequence of closed sets $C_n$ with $\operatorname{diam}C_n\to 0$ has nonempty intersection.
(quoted from S. Willard, General topology). If $\operatorname{diam}\varnothing =0$, the above is false without additional requirement that $C_n$ are nonempty.
That said, it's probably best to put nonempty there. The absence of nonempty leads to wrong statements in a number of books, e.g., "if $N$ is compact, there exist $x,y\in N$ such that $\rho(x,y)=\operatorname{diam}N$". (G.T. Whyburn, Analytic topology).
Summary.
- It's safer to keep $\operatorname{diam}$ nonnegative, because it may appear in formulas that need nonnegative inputs.
- If the validity of what you write depends on the interpretation of $\operatorname{diam}\varnothing$, consider changing the statement.
Best Answer
I'm not sure how much elaboration is needed, so I intend to be detailed.
Following @Randall s point, as you said every set in a discrete metric space every set is open, because of the reason you mentioned; and therefore every set is complement of an open set and therefore is closed. So $ \overline{A} = A $ and $\overline{X \setminus A} = X \setminus A$. The definition $ \partial A = \overline{X \setminus A} \cap \overline{A} $ is equivalent to yours (you can check this). And therefore for any $A$ in a discrete metric space we would have $ \partial A = A \cap (X \setminus A) = \emptyset $.
To use your definition more directly, here's another proof: Let $ x \in X$ be any candidate for being a member of $ \partial A $. Consider the open ball $ B(x, \frac{1}{2}) = \{ y \in X \mid \delta(x,y)<\frac{1}{2} \}$. For $x$ to be in $ \partial A $, it is necessary that $ B(x, \frac{1}{2}) \cap A \neq \emptyset $ and $ B(x, \frac{1}{2}) \cap (X \setminus A) \neq \emptyset$. But in the discrete metric space we know that $ B(x, \frac{1}{2}) = \{x\}$, because every distinct point from $x$ is 1 unit away from it. Now $x$ itself is either a member of $A$ or $X \setminus A$. In both cases the open ball $ B(x, \frac{1}{2}) $ does not contain a member of both $A$ and $ X \setminus A$. And therefore $x$ cannot be a member of $\partial A$. So the proof is accomplished.