Directly from the definition of product topology we know that for each $x\in\mathbb R$ there is a rational $\varepsilon_x>0$ such that $\{x\}\times(-\varepsilon_x,\varepsilon_x)\subseteq V$.
Now, for each $q\in\mathbb Q$ denote $A_q=\{x\in\mathbb R; \{x\}\times(-q,q)\subseteq V\}$. We have $\mathbb R=\bigcup_{q\in\mathbb Q} A_q$.
If every set $A_q$ would be finite, this would imply that $\mathbb R$ is countable, a contradiction.
(You could do basically the same thing working with the numbers of the form $\frac1n$, $n\in\mathbb N$, instead of rationals.)
You can also note that we would obtain the same contradiction by assuming that all $A_q$'s are countable, hence this shows that there is an $\varepsilon>0$ and an uncountable set $I$ with the property you're asking about.