I want to show that $S \neq \mathbb R^n$ and $S \neq \emptyset \implies \partial S \neq \emptyset$ using the following property : Let $S \subset \mathbb R^n$ a compact set and $T \subset \mathbb R^n$ a closed set, both are not empty et $S \cap T = \emptyset$. Then $\exists a \in S$ and $b \in T$ such that $$inf(\lvert\lvert x-y \rvert \rvert : x \in S, y \in T) = \lvert\lvert a-b \rvert \rvert > 0$$ I hesitate about the way to do this, I don't know what I need to take to be my compact set and closed set and the next steps are also not very clear.
(I already know the easy proof with $\partial S = \emptyset \implies \overline{S}=\mathring{S}=S \implies S = \mathbb R^n$ or $S = \emptyset $, I want to use the given property to show the proposition.)
Let $S \ne \emptyset, \Bbb{R}^n$ be a set such that $\partial S = \emptyset$. Then there exists a closed ball $K$ large enough such that $\overline{S} \cap K$ and $\overline{S^c} \cap K$ are both nonempty. Hence they are two disjoint nonempty compact sets so there exist $$a \in \overline{S} \cap K, \quad b \in \overline{S^c} \cap K$$ which minimize the distance between $\overline{S} \cap K$ and $\overline{S^c} \cap K$.
Consider the element $\frac{a+b}2$. It is certainly in $K$ since $K$ is convex. We have $$\left\|a-\frac{a+b}2\right\| = \left\|b-\frac{a+b}2\right\| = \frac12\|a-b\|$$ so
$\frac{a+b}2$ is closer to $a$ than $b$ is and hence it cannot be in $\overline{S^c} \cap K$,
$\frac{a+b}2$ is closer to $b$ than $a$ is and hence it cannot be in $\overline{S} \cap K$.
This is a contradiction since clearly $$K = (\overline{S} \cap K) \cup (\overline{S^c} \cap K).$$