Non-Archimedean Division Ring is Totally Disconnected

Theorem
Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring.

Let $\tau$ be the topology induced by the norm $\norm{\,\cdot\,}$.

Then the topological space $\struct {R, \tau}$ is totally disconnected.

Proof
Let $S$ be a subset of $R$ such that:
 * $\exists x, y \in S: x \ne y$

Let $r \in \R_{>0} : r = \norm {x - y}$

Consider the open ball $\map {B_r} x$ such that:
 * $x \in \map {B_r} x$
 * $y \notin \map {B_r} x$

By Open Balls are Clopen then $\map {B_r} x$ is both open and closed.

By Complement of Clopen Set is Clopen then $R \setminus \map {B_r} x$ is open.

Let:
 * $A = S \cap \map {B_r} x$

and
 * $B = S \cap \paren{R \setminus \map {B_r} x}$

We have:
 * $x \in A \subseteq \map {B_r} x$

and
 * $y \in B \subseteq \paren{R \setminus \map {B_r} x}$

By definition of separated sets:
 * $A$ and $B$ are separated sets

As $S = A \cup B$ then:
 * $S$ is the union of two non-empty separated sets of $\struct{R,\tau}$

By definition of connected subset:
 * $S$ is not a connected subset.

Since $S$ was an arbitrary subset of $\struct{R,\tau}$ with two distinct elements:
 * no set $S$ with two distinct elements is connected.

From Singleton is Connected in Topological Space:
 * the only connected subsets of $\struct{R,\tau}$ are the singletons

By definition of totally disconnected:
 * $\struct{R,\tau}$ is totally disconnected