Infinite product of connected spaces may not be connected? Let $X$ be a connected topologoical space. Is it true that the countable product $X^\omega$ of $X$ with itself (under the product topology) need not be connected? I have heard that setting $X = \mathbb R$ gives an example of this phenomenon. If so, how can I prove that $\mathbb R^\omega$ is not connected? Do we get different results if $X^\omega$ instead has the box topology?

The first part of your question - about connectedness of $\mathbb
R^\omega$ with the usual product
topology - has
already been answered.
To show that box
product $\mathbb
R^\omega$ is not connected we only need find a clopen subset $U$ of this
topological space (different from $\emptyset$ and the whole space). 
Here are two examples of such sets:


*

*$U=$ set of all sequences, that converge to $0$; as suggested by Henning's
comment, see also here
here.
Indeed, if $x_n\to 0$ then $V=\prod(x_n-1/2^n,x_n+1/2^n)$ is a
neighborhood of $x$ such that $V\subseteq U$, therefore $U$ is
open. Similar argument shows that the complement of $U$ is open in
the box topology.

*$U=$ set of all sequences that are bounded; see e.g. Example
10.16
here.
The argument is similar, here we can even use open intervals of
the same length on each coordinate.
