Identify the boundary of $\mathbb{Q}$ in the discrete metric space $\mathbb{R}$

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.