Argh! Suppose it is disconnected, so that there is a set $A$ which is both open and closed. For any sequence $\mathbf{x}\in A$ , there must be $B=B(\mathbf{x},\epsilon)\subset A$ , and for any $\mathbf{y}\notin A$ , there must be $B'=B(\mathbf{y},\delta)\subset\mathbb{R}^{\omega}-A$ . The set $B$ contains all sequences that differ from $\mathbf{x}$ by no more than $\delta<\epsilon$ in each coordinate, which still have to be in $A$ . And similarly, for $B'$ , $\mathbf{y}$ and $\mathbb{R}^{\omega}-A$ . So, for example, $\overline{\mathbb{R}^{\infty}}$ in the uniform topology, i.e. the set of sequences converging to 0, would not work. At the same time, if $A$ is the set of bounded sequences, and $X-A$ is the set of unbounded sequences, then both seem to be open in the uniform topology. So, the answer is no, it is not connected.