Assume $E\neq \emptyset $, $E \neq \mathbb{R}^n $. Then prove $E$ has at least one boundary point. (i.e $\partial E \neq \emptyset $).
=================
Here is what I tried.
Consider $P_0=(x_1,x_2,\dots,x_n)\in E,P_1=(y_1,y_2,\dots,y_n)\notin E $.
Denote $P_t=(ty_1+(1-t)x_1,ty_2+(1-t)x_2,\dots,ty_n+(1-t)x_n) $, $0\le t\le 1$.
$ t_0=\sup\{t |P_t \in E\} $. And then I wanted to prove that $P_{t_0}\in \partial E$.
A. If $P_{t_0}\in E$. then $t\neq 1$ otherwise $P_{t_0}=P_1$. And by definition ,$P_t \notin E$ for $t_0 \lt t \leq 1 $. And choose $t_n$,such that $1\gt t_n\gt t_0$,$t_n \to t_0$, which makes $P_{t_n} \notin E$, but $P_{t_n} \to P_{t_0}$.Then $P_{t_o} \in \partial E$.
B. If $P_{t_0}\notin E$. then $t\neq 0$ otherwise $P_{t_0}=P_0$.And then choose $t_n$ such that $0\lt t_n\lt t_0$ , $t_n \to t_0$ ,therefore $P_{t_n} \to P_{t_0}$ and $P_{t_n} \in E$. Hence $P_{t_o} \in \partial E$.
Thus we have $\partial E \neq \emptyset$.
Am I correct? the construct of $P_t$ is a clue from my elder. What I am wondering is this step in A(Also, B). $$P_{t_n} \to P_{t_0} \Rightarrow P_{t_o} \in \partial E$$ I can somewhat image this. But how to make this step strictly?
Here's a proof that doesn't use connectedness at all. Suppose that $\emptyset \neq E \subsetneq \mathbb{R}^n$. Now take $x \in \mathbb{R}^n \setminus E$. If $x$ is a boundary point of $E$, we're done! Otherwise, take $\delta = \sup \{\epsilon : \epsilon > 0, B_{\epsilon}(x) \cap E = \emptyset \}$, where $B_\epsilon(x)$ is the open ball surrounding $x$ with radius $\epsilon$. Now since $x$ is not a boundary point and $E$ is non-empty, we know that this $\delta$ exists and is well defined.
Take $S = \{s \in \mathbb{R}^n : |s - x| = \delta \}$. That is, $S$ is the boundary of the open ball centered at $x$ with radius $\delta$. I claim that there is a point in $S$ which is on the boundary of $E$.
Define the following function: $f:S \rightarrow \mathbb{R}$ such that $f(s) = \inf_{e \in E} |s - e|$. There exists some $\hat{s} \in S$ such that $f(\hat{s}) = 0$. If there is no such $\hat{s}$, our choice of $\delta$ was not maximal, which would contradict the definition of $\delta$.
EDIT: It was pointed out that this part of the argument requires$f$ to be continuous. If you already have the fact that the distance function between a point and a set is continuous, you can ignore this part of the post. To prove that fact, let $(s_n) \rightarrow s$ be a convergent sequence in $S$.
Observe by the triangle inequality that for any $e \in E$, we have: $$\begin{align*} f(s_n) & \leq |s_n - e| \\ & \leq |s_n - s| + |s - e| \end{align*} $$
Taking sufficiently large $n$ we can get $|s_n - s| < \epsilon$. Additionally, taking an infimum over our choice of $e$ yeilds: $$f(s_n) \leq f(s) + \epsilon$$
On the flip side, we get for any $e \in E$: $$\begin{align*} f(s) & \leq |s - e| \\ & \leq |s_n - s| + |s_n - e| \\ & \leq \epsilon + |s_n - e| \end{align*} $$
taking an infimum over $e \in E$ yields: $$f(s_n) \geq f(s) - \epsilon$$ Combining our inequalities: $$|f(s) - f(s_n)| \leq \epsilon$$ This proves the continuity of $f$
Now take $\epsilon > 0$, and the open ball $B_\epsilon(\hat{s})$. This open ball contains a point of $E$ and a point of $E^c$. To see why, notice that since $f(\hat{s}) = 0$, there is some point $e \in E$ such that $|\hat{s} - e| < \epsilon$. Further, since $\hat{s}$ is in the set $S$, we have that $B_\epsilon(\hat{s}) \cap B_{\delta} (x) \neq \emptyset$. and since $B_{\delta} (x) \subset E^c$, we are done.
Therefore, $\hat{s}$ satisfies the definition of a boundary point of $E$.
Note that there is a better way to prove this statement using the connectedness of $\mathbb{R}^n$, but the comments indicate that the OP does not want to use this connectedness.