Consider $R^{\omega}$ in the box topology.
x & y are in the same component iff $x_i = y_i$ for infinitely many values.
I proved the --< implication, but I can't come up with the proof for the other direction. I was thinking to do it by contradiction that is suppose x & y are in the same components and $x_i \neq y_i$ for infinitely many values, but I couldn't proceed further. The problem that I am having is that in order to show this I have to show there doesn't exist any connected subspace of $R^{\omega}$ containing both x and y.
proof:
Suppose that x and y are in same component and $x_i = y_i$ for finitely many values of $i \in \mathbb{Z^{+}}$.
Let $\phi_i : R \rightarrow R$ be a homeomorphism that depend on i such that $\phi_i(x_i) = 0$ and $\phi_i(y_i) = i$ when $y_i \neq x_i$.
Consider $\Pi\ f_i = \tau\ : \ R^{\omega} \rightarrow R^{\omega}$ this is a homeomorphism in $R^{\omega}$ with the box topology (trivial to proof).
Now here is where I am having troubles;I can see intuitively that we will have connected component U and V that separate $\tau(x)$ and $\tau(y)$ but I can't prove that rigorously if someone could help in that final step that would be great.
This was originally published as part of this note which covered all three Munkres topologies on $\mathbb{R}^\omega$ (based on the posting on topology atlas forums. For easier reference on this site I re-use the most complicated part on the box topology:
Let $X = \mathbb{R}^\omega$ be a countable product of copies of $\mathbb{R}$., and we give $X$ the box-topology.
Define a relation $\sim$ on $X$ as follows: $x \sim y$ iff the sequence $x_n - y_n$ is $0$ from a certain index onwards (or equivalently if the set of $\{n: x_n \ne y_n\}$ is finite). This is an equivalence relation: $x \sim x$ because then we have a $0$-sequence, symmetry is evident, as $-0 = 0$, and if $N_1$ is an index from which $x_n - y_n = 0$, and $N_2$ a similar one for $y_n - z_n$, then $\max\{N_1,N_2\}$ works for $x_n$ and $z_n$.
I'll show first that the classes $[x]$ are path-connected. So fix $x$. Then define for each finite subset $I$ of $\omega$ the set $X(I) = \{(y_n) : y_n = x_n \text{ for all $n$ in $\omega \setminus I$}\}$. This is homeomorphic to $\prod_{n \in I} \mathbb{R}$ by the obvious map (and by noting that on a finite(!) product the box topology and the normal product topology agree, and a finite (normal) product of copies of $\mathbb{R}$ is path-connected). Also for all such $I$, $x$ is in $X(I)$. And $[x] = \bigcup\{X(I) : I \subset N, \text{ $I$ finite}\}$, by definition. And this union is path-connected because we can always connect two points in different $X(I)$'s via their common point $x$.
Claim (to finish): if $p$ is a point not in $[x]$, so it differs with $x_n$ for infinitely many $n$, then there is a subset $Y$ such that $p \in Y$ and $Y$ clopen (i.e. closed and open) and $x \notin Y$.
This shows that no connected subset can contain both $x$ and $y$, so $[x]$ is a maximal connected subset and hence a (path-)component of $X$.
To this end: let $U_{k,n}$ be open subsets of $\mathbb{R}$ (in the $n$-th component space) such that
It is obvious such sets can be chosen. Define $$ Y = \{(y_n) : \text{there is a $k$ such that for infinitely many $n$ we have $y_n \not\in U_{k,n}$}\}. $$ This $Y$ does not contain $x$, as then for all $k$,$n$ we have $x \in U_{k,n}$. It does contain $p$, as $p$ differs on infinitely many places from $x$, so for every $k$ there are infinitely many $n$ such that $p_n \not\in U_{k,n}$, by (2) above.
It remains to show that $Y$ is closed and open.
$Y$ is open: let $z$ be in $Y$. Let $k$ be given as in the definition of $Y$. So there are infinitely many indices $n$ (say $n$ in $N'$) such that $z_n$ is not in $U_{k,n}$. In particular, such $z_n$ are in the complement of $U_{k+1,n}$. Then the box-topology basic open subset $\prod_{n \in \omega} O_n$ where $O_n$ is the complement $U_{k+1,n}$, if $n$ in $N'$ and $O_n$ is $\mathbb{R}$ otherwise, is contained in $Y$ and contains $z$. (It is contained in $Y$ because $k+1$ works in the definition for $Y$ for all points of this open set.)
Y is closed: let $z \not\in Y$. So for all $k$ there are only finitely many $n$ such that $z_n \not\in U_{k,n}$. Call these finite exception sets $I_k$. These sets are increasing: if $n$ is in $I_m$ then also in $I_{m+1}$. If $n$ is in neither of the $I_k$, $z_n$ must be in the intersection (over $k$) of the $U_{k,n}$. Let $f$ be some strictly increasing map from $\omega \setminus \bigcup I_k$ into $\omega$ (if this set is non-empty). Define the following open set $O = \prod_{n \in \omega} O_n$ in the box topology: if $n$ is in $I_{k+1} \setminus I_k$, for some $k \ge 0$, let $O_n$ be $U_{k,n} \setminus \operatorname{cl} U_{k+1,n}$, for $n$ in $I_0$, let $O_n$ be $\mathbb{R}$, and for $n$ in $\omega \setminus U_{k,n}$.
I claim that $O$ contains $z$ (this is quite clear), and that $O$ is disjoint from $Y$. Why is this so? Let $w$ be in $O$. Pick $k$ in $\omega$. If $w_n$ is not in $U_{k,n}$, then we want to show there can be only finitely many of these $n$. Consider the cases for $n$: $n$ can be in $I_{k+1} \setminus I_k$ only if $i+1 \le k$, so there are only finitely many $k$ that can apply. And the union of these is finite. If $n$ is in $I_0$, who cares? Only finitely many $n$ are in $I_0$. And finally, if $n$ is in $\omega \setminus \bigcup I_k$, this can only be a problem if this is set is infinite, but then $f$ grows larger than $k$ from some index $N$ onwards, and then $w_n$ is in $U_{f(n),n} \subset U_{k,n}$ (remember that the $U$'s decrease in $k$) for $n \ge N$. So in $\omega \setminus \bigcup I_k$ there can also only be finitely many $n$ with $w_n \not\in U_{k,n}$. So this shows that for every $k$, all but finitely many $n$ have $w_n \in U_{k,n}$. So $w$ is not in $Y$, and $O$ is disjoint from $Y$. So $Y$ is closed.