I want to prove that: In a topological vector space $E$ a set different from $\emptyset$ and from $E$ cannot be both open and closed, in the other words, if a subset $M\subset E$ is open and closed, then $M= \emptyset$ or $M=E$.
I can prove that $M \subset E$ is a linear subspace of $E$ then, $M=E$. But, in our case, $M$ is a general set (closed and open).
Any topological vector space $X$ is path connected: Given $x$ and $y$ the map $f:[0,1] \to X$ defined by $f(t)=tx+(1-t)y$ is a path from $y$ to $x$. Any path connected space is connected. If $E$ is open and closed that $X=E \cup E^{c}$ gives a disconnection of $X$.
Direct proof without using path-connectedness:
Suppose such a set $E$ exists. Pick $x \in E$ and $y \in E^{c}$. Let $f(t)=tx+(1-t)y$. Consider $c=\sup \{t: f(t) \in E^{c}\}$. There exist $t_n$ increasing to $c$ such that $f(t_n) \in E^{c}$ for all $n$. Since $f$ is continuous and $E^{c}$ is closed we get $f(c) \in E^{c}$. Similarly, taking a sequence $t_n$ decreasing to $c$ we get $f(c) \in E$. [Note that $c \neq 1$ since $f(1)=x \in E$]. Thus $f(c) \in E \cap E^{c}=\emptyset$, a contradiction.