Your partial solution is correct as far as it goes. Path connectedness, like connectedness, is productive (see here or here), so $\Bbb R^\omega$ is also its only component in the product topology.
Identifying the components of $\Bbb R^\omega$ in the uniform topology is a little tricky. It’s easiest to start with an arbitrary $x=\langle x_n:n\in\omega\rangle\in\Bbb R^\omega$ and describe the component $C(x)$ containing $x$. In fact, it’s easiest to start with $C(z)$, where $z=\langle 0,0,0,\ldots\rangle$ is the point with all coordinates $0$. Let $B$ be the set of bounded sequences in $\Bbb R^\omega$; you already know that $B$ is clopen in the uniform topology, and I claim that it’s also connected and hence that $C(z)=B$.
Let $x=\langle x_n:n\in\omega\rangle\in B\setminus\{z\}$. Define
$$f:[0,1]\to\Bbb R^\omega:t\mapsto\langle tx_n:n\in\omega\rangle\;,$$
and note that $f(0)=z$ and $f(1)=x$. Let $t\in[0,1]$ and $\epsilon>0$ be arbitrary, and let $N$ be the $\epsilon$-ball centred at $f(t)$:
$$N=\{\langle y_n:n\in\omega\rangle\in\Bbb R^\omega:|y_n-tx_n|<\epsilon\text{ for all }n\in\omega\}\;;$$
what condition on $s\in[0,1]$ will ensure that $f(s)\in N$, i.e., that $|sx_n-tx_n|<\epsilon$ for all $n\in\omega$? Clearly we want to have $|s-t|<\frac{\epsilon}{|x_n|}$ for all $n\in\omega$. Is this possible? Yes, because $x\in B$, and therefore $\{|x_n|:n\in\omega\}$ is bounded. Let $\|x\|=\sup_n|x_n|$, and let $\delta=\frac{\epsilon}{\|x\|}$; then $f(s)\in N$ whenever $|s-t|<\delta$, and $f$ is therefore continuous. Thus, $B$ is even path connected and is both the component and the path component of $z$.
Now let $x\in\Bbb R^\omega$ be arbitrary; then $C(x)=B+x$, where as usual $B+x=\{y+x:y\in B\}$, and $C(x)$ is also the path component of $x$. To see this, simply note that the map $$f_x:\Bbb R^\omega\to\Bbb R^\omega:y\mapsto y+x$$ is a homeomorphism, that $x=f_x(z)$, and that $B+x=f_x[B]$. It’s not hard to see that we can also describe $C(x)$ as the set of $y\in\Bbb R^\omega$ such that $y-x$ is bounded.
Corrected Version (8 February 2015):
Finally, we consider $\Bbb R^\omega$ with the box topology. This is a significantly harder problem. For $x\in\Bbb R^\omega$ let $$F(x)=\big\{y\in\Bbb R^\omega:\{n\in\omega:x_n\ne y_n\}\text{ is finite}\big\}\;;$$ I’ll show first that $F(x)$ contains the component of $x$.
Suppose that $y\in\Bbb R^\omega\setminus F(x)$. For $n\in\omega$ let $\epsilon_n=|x_n-y_n|$. If $\epsilon_n>0$ and $k\in\omega$ let
$$B_n(k)=\left(x_n-\frac{\epsilon_n}{2^k},x_n+\frac{\epsilon_n}{2^k}\right)\subseteq\Bbb R\;,$$
and note that $y_n\notin B_n(k)$ for any $k\in\omega$. For each $k\in\omega$ let
$$U_k=\big\{z\in\Bbb R^\omega:\{n\in\omega:z_n\notin B_n(k)\}\text{ is infinite}\big\}\;,$$
and let $U=\bigcup_{k\in\omega}U_k$. Clearly $y\in U$ and $x\notin U$.
Let $z\in U$; then $z\in U_k$ for some $k\in\omega$. Let $M=\{n\in\omega:z_n\notin B_n(k)\}$. Then $$z_n\notin B_n(k)\supseteq\operatorname{cl}_{\Bbb R}B_n(k+1)$$ for each $n\in M$. For $n\in M$ let $G_n=\Bbb R\setminus\operatorname{cl}_{\Bbb R}B_n(k+1)$, and for $n\in\omega\setminus M$ let $G_n=\Bbb R$; then $G=\prod_{n\in\omega}G_n$ is open, and $z\in G\subseteq U_{k+1}\subseteq U$, so $U$ is open.
Now suppose that $z\in\Bbb R^\omega\setminus U$; then $\{n\in\omega:z_n\notin B_n(k)\}$ is finite for each $k\in\omega$. In other words, for each $k\in\omega$ there is an $m_k\in\omega$ such that $z_n\in B_n(k)$ for all $n\ge m_k$. Clearly we may assume that $m_k<m_{k+1}$ for each $k\in\omega$. For $m_k\le n<m_{k+1}$ let $G_n=B_n(k)$, and for $n<m_0$ let $G_n=\Bbb R$. Then $G=\prod_{n\in\omega}G_n$ is open, and $z\in G\subseteq\Bbb R^\omega\setminus U$, so $U$ is closed.
Thus, $U$ is a clopen set separating $x$ and $y$, and the component of $x$ must therefore be a subset of $F(x)$.
In fact $F(x)$ is the component and path component of $x$. To see this, suppose that $y\in F(x)$, and let $D=\{n\in\omega:x_n\ne y_n\}$, so that $D$ is finite. Let
$$Y=\{z\in\Bbb R^\omega:z_n=x_n\text{ for all }n\in\omega\setminus D\}\;;$$
then $Y$ is homeomorphic to $\Bbb R^{|D|}$. $\Bbb R^{|D|}$ is path connected, and a path in $\Bbb R^{|D|}$ transfers easily to a path in $Y$ and hence in $\Bbb R^\omega$ via the homeomorphism.
Best Answer
The set $U$ only contains bounded sequences if $(a_n)$ is bounded, so is a subset of $A$, and that's all you need because it shows that any bounded sequence is an interior point of $A$. We don't need it to contain all bounded sequences: just an open set around $(a_n)$ inside $A$.
The same holds for $B$: if $(a_n)$ is unbounded, the defined $U$ also contains only unbounded sequences as well, so $U \subseteq B$ and the sequence is an interior point of $B$.
So $A$ and $B$ only have interior points so both are open, and as $A \cup B = \Bbb R^\omega$, both are closed too, and we have a disconnection.