Given (X, ||•||) normed space, prove that only X itself and empty space are clopen. I' d like to ask you for some help. I' ve to prove the problem stated in title, but without using the knowledge that normed space is connected.And I just got no idea how to do so...
Thanks for any advice
 A: Another idea, if you can use that $\Bbb R$ is connected: the space $X$ is a union of connected subsets with nonempty intersection, namely the rays $\{tx\,\vert\,t\in\Bbb R\}$, $x\in X$, $x\ne 0$ because each ray is a continuous image of $\Bbb R$.
A: A normed space is normal. Suppose that $U\subset X$ with $\emptyset\neq U\neq X$ is a clopen set. In particular, $U$ and $U^c$ are closed. Since $(X,\|\cdot \|)$ is normal, there is $A,B$ open such that $U\subset A$, $U^c\subset B$ and $A\cap B=\emptyset$. But this is impossible ! (why ?)
Hint : Define $$d(X,Y)=\inf\{\|x-y\|\mid x\in X, y\in Y\}$$
and use the fact that $d(U,U^c)=0$ to show that there is at least one element in $A\cap B$.
A: Suppose that exists $U\neq \varnothing, X$ so that both $U$ and $X\backslash U$ are closed. Take $x\in U$ and $y\in X\backslash U$. 
Denote by $I_0$ the line segment connecting $x$ and $y$. Divide $I_0$ into two sub-segments by the midpoint. One of those must have endpoints in both $U$ and $X\backslash U$, call this segment $I_1$.
Keep dividing as above, we obtain a sequence of shrinking segments
$$I_0\supset I_1\supset I_2\supset \cdots$$
These segments share a commmon point $z$, which set would $z$ belong to: $U$ or $X\backslash U$? Neither: as $z$ is the limit of both of the end-points sequences.
