So if I'm reading correctly you want to find out if there is a continuous (with respect product topology) surjective map $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$?
No, there is not. Note that $\mathbb{R}$ is $\sigma$-compact, so write:
$\mathbb{R} = \bigcup_{n \in \mathbb{N}} [-n,n]$
Then using the fact that $f$ is surjective we get:
$\mathbb{R}^{\omega} = \bigcup_{n \in \mathbb{N}} f([-n,n])$
By continuity of $f$ each $D_n=f([-n,n])$ is a compact subset of $\mathbb{R}^{\omega}$. So the question boils down to whether is possible that $\mathbb{R}^{\omega}$ is $\sigma$-compact with product topology.
No, let $\pi_{n}$ be the standard projection from $\mathbb{R}^{\omega}$ onto $\mathbb{R}$, then $\pi_{n}f([-n,n])$ is a compact subset of $\mathbb{R}$ so bounded. Thus for each $n \in \mathbb{N}$ choose $x_{n} \in \mathbb{R} \setminus \pi_{n}f([-n,n])$ then $x=(x_{n})$ lies in $\mathbb{R}^{\omega}$ but not in $\bigcup_{n \in \mathbb{N}} f([-n,n])$, a contradiction.