Totally disconnected metric space I want to prove that a certain metric space is totally disconnected.
In a metric space context this is the same as saying that every connected component is a singleton. 
I think another way of proving that a space is TD is proving that there is a proper, nonempty open and closed set. Is that right?
Please let me know any alternative equivalent definitions you might now.
Cheers!
 A: The following two facts concerning totally disconnected spaces should (separately) help you demonstrate that your space is totally disconnected.
Fact 1: Every T$_1$-space with a basis consisting of clopen sets (i.e., every zero-dimensional space) is totally disconnected.
proof. Suppose that $A \subseteq X$ contains at least two points, and let $x , y \in A$ be distinct.  By assumption there is a clopen $U \subseteq X$ such that $x \in U$ and $y \notin U$.  But then $U$ and $X \setminus U$ witness that $A$ is not a connected subset of $X$. $\quad\Box$
Fact 2: Every product of (nonempty) totally disconnected spaces is totally disconnected.
proof. Suppose that $X_i$ is totally disconnected for all $i \in I$, and let $A \subseteq \prod_{i \in I} X_i$ contain at least two points.  Then there must be a $j \in I$ such that $A_j = \{ x_j : x = ( x_i )_{i \in I} \in A \}$ contains at least two points.  As $X_j$ is totally disconnected, there are open $U_j , V_j \subseteq X_j$ such that $U_j \cap A_j \neq \emptyset \neq V_j \cap A_j$ and $U_j \cap V_j \cap A_j = \emptyset$ and $A_j \subseteq U_j \cup V_j$.  Let 
$$U = {\textstyle \prod_{i \neq j}} X_i \times U_j; \quad
V = {\textstyle \prod_{i \neq j}} X_i \times V_j.$$
Then $U , V$ are open subsets of $\prod_{i \in I} X_i$, $U \cap A \neq \emptyset \neq V \cap A$, $U \cap V \cap A = \emptyset$ and $A \subseteq U \cup V$.  Thus $A$ is not a connected subset of $\prod_{i \in I} X_i$. $\quad\Box$
Either of these should be useful (but especially the second) because your space appears to be a subspace of $\{ 1 , \ldots , k \}^{\mathbb{N}}$ taking $\{ 1 , \ldots , k \}$ to be discrete, and then taking the product topology.  (Also, total disconnectedness is a hereditary property of topological spaces.)
A: Instead of totally disconnected, your definition in fact is that of disconnected, and your equivalent definition is correct. A space is connected if and only if the only closed and open sets are the empty set and the whole space.
