"Proof" that $\mathbb{R}^J$ is not normal when $J$ is uncountable

Even without seeing just where the problem is, you can see that there must be one: if $m \ne n$, then $X\setminus P_m$ is an open nbhd of $P_n$ properly contained in $X$, so your conclusion that there can be no such set is false.

The flaw in your argument is that the open nbhd $U$ of $P_n$ need not be a basic open nbhd: it need not be a product of open sets. In fact, no set of the form $U(x,B)$ can contain $P_n$. To see this, let $B_0 = \{\alpha\in B:x(\alpha)=n\}$, and let $B_1 = B \setminus B_0$. If $B_1 \ne \varnothing$, define $y \in X$ by $y(\alpha) = n$ for all $\alpha \in J$; clearly $y \in P_n \setminus U(x,B)$. If $B_1 = \varnothing$, fix $\alpha_0 \in B_0$, and define $y \in X$ by $$y(\alpha) = \begin{cases}n+1,&\alpha = \alpha_0\\ n,&\text{otherwise}; \end{cases}$$ clearly we again have $y \in P_n \setminus U(x,B)$.

You cannot prove what you claim: $P_1$ and $P_2$ are closed and disjoint so clearly $X \setminus P_2$ is an open superset of $P_1$, unequal to $X$.

The mistake in the proof is that you show: for every coordinate $\alpha$ and every $m$ in $\mathbb{Z}^{+}$ there is a basic neighborhood $U(\alpha,m)$ that it is a subset of $U$ and $m \in p_{\alpha}[U(\alpha, m)]$. It does not follow from that that $U = X$ from that. It only shows that $U$ projects onto each factor space, but this is not enough. Consider the diagonal in a finite product of discrete spaces, for an example of that: the diagonal is open, and projects onto each factor in the full set, yet is "thin" as well, and not equal to the full product.