The relation between a metric space $(X,d)$ and the topological space that arises from it. Consider the topological space $(\Bbb R,\mathfrak I)$ that arises from the metric space $(\Bbb R,d)$, with $d(x,y)=|x-y|$. I want to prove that $\partial(a,b)=\partial[a,b]=\{a,b\}$. 
I have that 
$$x\in \partial A \iff d(x,A)=0\wedge d(x,X\setminus A)=0$$
[This used to be a much longer and tortuous question, but since I can't delete it, I'll just leave what might interest other users, though it wasn't my main concern. When I find a suitable way to ask about my concern, I'll edit]
 A: If you want to use
$$\partial A = \overline{A}\cap \overline{X-A}$$
you can proceed as follows:


*

*Step 1. If $x\in (a,b)$, then $x\notin\partial{(a,b)}$ and $x\notin\partial{[a,b]}$.
Proof. Let $\epsilon = \min\{x-a,b-x\}$. Then $\{ y\mid d(x,y)\lt\frac{\epsilon}{2}\}\subseteq (a,b)$ and is open, so $x\notin\overline{X-(a,b)}$; since $\overline{X-[a,b]}\subseteq\overline{X-(a,b)}$, it follows that $x\notin\overline{X-[a,b]}$.

*Step 2. If $x\in (-\infty,a)$ then $x\notin\partial{(a,b)}$ and $x\notin\partial[a,b]$.
Proof. Let $\epsilon =a-x$. Then $\{y\mid d(x,y)\lt \frac{\epsilon}{2}\}\subseteq (-\infty,a)\subseteq X-[a,b]$ and is open, so $x\notin\overline{[a,b]}$. Since $\overline{(a,b)}\subseteq\overline{[a,b]}$, it follows that $x\notin\overline{[a,b]}$.

*Step 3. If $x\in (b,\infty)$, then $x\notin\partial{(a,b)}$ and $x\notin\partial[a,b]$.
Proof. Similar to that in step 2.

*Step 4. $a\in\partial(a,b)\cap\partial[a,b]$.
Proof. Let $\epsilon\gt 0$. Let $\delta=\frac{1}{2}\min\{b-a,\epsilon\}$. Then $a+\delta\in (a,b)\cap \{x\mid d(x,a)\lt\epsilon\}$, and $a-\delta\in (X-[a,b])\cap \{x\mid d(x,a)\lt\epsilon\}$. Thus, every open ball containing $a$ intersects $(a,b)$, so $a\in\overline{(a,b)}\subseteq \overline{[a,b]}$; and every open ball containing $a$ intersects $X-[a,b]$ and so $a\in\overline{X-[a,b]}\subseteq\overline{X-(a,b)}$. Thus, $a\in \partial (a,b)\cap\partial[a,b]$.

*Step 5. $b\in\partial(a,b)\cap\partial[a,b]$.
Proof. Similar to step 4. 
