Uncountable Cartesian product of closed interval

It is not.

Let's look at open sets containing 0 (sequence of 0s).

We will argue by contradiction, so suppose there is a countable local neighborhood basis $U_i$ for 0. Every such $U_i$ contains some $V_i = \prod_{r} V_{i,r}$ where $V_{i,r} = I$ for almost every $r$, $V_{i,r}$ open. (By definition of product topology).

Let's look at the set of all $r$ that $V_{i,r} \neq I$ for some $i$.

This set is countable, because it's a countable union of countable sets. So it's not the whole of $I$. Let's choose some $r_0$ outside this set.

Let $H = \prod_r H_r$ where $H_{r_0} = [0, 1/2)$ and $H_{r} = I$ otherwise.

Then $H$ is an open set containing $0$, not containing any of $U_i$, contradicting first-countability at 0.