Infinite product is not locally compact. I have read that a space $X$ at a point $x$ Is locally compact if there is a compact subspace $C$ of $X$ that contains a nbd of x.
Now to show that infinite product $R^w$  is not locally compact let $0=(0)_{n=1}^{\infty}$ 
And nbd be $(-e,e)*(-e,e)*...*(-e,e)*R*R....$, it is not locally compact (otherwise closure of the nbd would be compact).
But if we replace the nbd by $(-e,e)*(-e,e)*...*(-e,e)*...$. Can we say that it is locally compact.
Note : $(-e,e)$ is e-nbd of 0. 
 A: In his answer I show that whenever $\prod_{i \in I} X_i$ is locally compact, all $X_i$ are locally compact and all but finitely many of the $X_i$ are compact.
Your case has $X_n = \mathbb{R}$ for all $n \in \mathbb{N}$ so no compact factors at all and infinitely many spaces, so the product is not locally compact. 
The flaw in your argument is indeed, as mentioned in a comment, that $(-\varepsilon, +\varepsilon) \times (-\varepsilon, +\varepsilon) \times \ldots$ is not a neighbourhood of the all $0$-point in the product topology, as it does not contain a basic open set of the form $\prod_{i \in I} O_n$ where all $O_n$ are open and all but finitely many $O_n$ are equal to $\mathbb{R}$. It's a common confusion, if we want your set to be open we need a finer topology, the box topology, but in that topology $\mathbb{R}^\omega$ is not locally compact either, so no improvement. 
Note that $\mathbb{R}^\omega$ is an infinite-dimensional topological vector space and these are never locally compact by a classic theorem that says that a locally compact TVS is finite-dimensional.
