connectedness of uniform topology This question can also be found in this link, but because of dull brain I could not understood that One line proof.

Let $\mathbb{R}^{\omega}$ be the set of all (infinite) sequences of real numbers. Determine whether $\mathbb{R}^{\omega}$ is connected or not in the uniform topology.
Basis elements of uniform topology are $B_{\rho}(x,\epsilon)=\{ y\mid \rho(x,y)<\epsilon \}$, where $\rho(x,y)=\sup\{ \min(d(x_\alpha,y_\alpha),1)\mid \alpha \in \mathbb{Z} \}$

From the answer in the link, I understood that it is not connected.
What I found:
Note that $B_{\rho}(x,1)=\overline{B_{\rho}(x,1)}$. So I need to find $x_1,x_2\in \mathbb{R}^{\omega}$ such that $B_{\rho}(x_1,1)\cap B_{\rho}(x_2,1)=\phi$
How to find such $x_1$ and $x_2$?
 A: So we consider on $\mathbb{R}^\omega$ what Munkres calls the uniform topology, i.e. the topology induced by the metric 
 $$d(x,y) = d((x_n)_n,(y_n)_n) = \sup_n \{\min(|x_n - y_n|,1): n \in \mathbb{N}\}$$
Define the following relation $\sim$ on $X$: $x \sim y$ iff the sequence $(x_n - y_n)_n$ is bounded in $\mathbb{R}$.
This is an equivalence relation on $X$: 


*

*$x \sim x$ because the $0$-sequence is bounded.

*$x \sim y$ implies $y \sim x$ because $(x_n -y_n)_n$ is bounded iff 
its negative $(y_n - x_n)_n$ is.

*$x \sim y$ and $y \sim z$ imply $x \sim z$ because if the sequences 
$(x_n - y_n)_n$ and $(y_n - z_n)_n$ are bounded then their sum 
$(x_n -z_n)_n$ is bounded too.
So, we have a partition into equivalence classes. 
Denote the equivalence class of $x=(x_n)\in \mathbb{R}^\omega$ by $[x]$.
Also note that all classes are mutually homeomorphic:
If we have a fixed $x \in \mathbb{R}^\omega$, then the map $f_x$ that sends $(y_n) \in [x]$ to
$(y_n - x_n)$ in $[0]$ (the class of the $0$-sequence, i.e, all 
bounded sequences in $X$) is a homeomorphism.
($X$ is a topological group under $+$, or check continuity 
with the definition; it's even an isometry.)
Also, every set $[x]$ is open in $X$. For, let $y \in [x]$.
Then the open ball $B_d(y, {1 \over 2}) \subset [x]$: if $z  \in B_d(y,{1 \over 2})$ this implies that for
all $n$,  $|y_n - z_n| < {1 \over 2}$, so $y_n - z_n$ is bounded, so $z$ is in $[y]=[x]$.
So we have a partition of open sets, so they are all closed as well, as the complement of a class is
the union of the other classes, hence also open.
Note also that $[0]$ is a topological vector space, it is homeomorphic to
$\ell_{\infty}$; or see in another way that $[0]$ (and hence all classes) are path-connected:
Every $x$ in $[0]$ is connected by a path $F_x:[0,1] \to [0]$ from $0$ to $x$:
$F(t) = ((t*x_n)$, for $t \in [0,1]$. This is continuous on $[0]$
because there $d(F_x(t_1),F_x(t_2)) = \sup_n \{\min(1, |t_1-t_2|*|x_n|)\}$;
now let $M$ be a bound on all the $|x_n|$ (by boundedness of $x$), and let $|t_1-t_2| < {1 \over M}$. Then: 
$d(F_x(t_1),F_x(t_2)) < M\cdot |t_1-t_2|$, which shows directly that $F_x$ is continuous.
So every point in $[0]$ is connected by a path to $0$, hence $[0]$ is path-connected, 
and as said, so are all other classes as well. Because the classes are clopen, there can be
no larger path-connected or connected set that has points in more than one class.
So sets of the form $[x]$ are maximal (path-)connected subsets.
So the $[x]$ ,$x \in X$ form the (path-)components of $X$.
As to your last part: take one sequence $x_n=0$, and the other $x_n =n$. Their balls of radius $\frac{1}{2}$ or $1$ do not intersect. These sequences clearly lie in different (and hence disjoint clopen) classes.
A: Not connected,  because there is set of bounded sequences that is both open and closed, in the uniform topology. 
If $A$ is this set of bounded sequences and $X - A$ be the set of unbounded sequence, then both seem to be open in uniform topology. 
So, that makes a separation on  $\mathbb{R}^\omega$ in the uniform topology. 
