The arbitrary product of Reals is Baire I'm slowly working my way through Munkres and need help with exercise 12 of section 48. The exercise reads:

Show that $\mathbb{R}^J$ is a Baire space in the box, product, and uniform topologies.

Here, Munkres' convention is that $J$ represents a potentially uncountable index set. In the uniform topology, Baireness of $\mathbb{R}^J$ follows immediately since for any metric space $X$, $X^J$ is complete under the uniform metric (and all complete metric spaces are Baire). However, the product topology is more difficult, and I haven't yet attempted the box toplogy. My proof so far for the product topology:
Let $\mathbb{R}^J$ be given the product topology and let $\{U_n\}$ be a countable collection of dense open subsets of $\mathbb{R}^J$. For any open subset $V$ of $\mathbb{R}^J$, there is some finite set $F \subset J$ such that $\pi_{\alpha_i}(V) \subset \mathbb{R}$ for every $\alpha_i \in F$. If $U_n$ is dense in $\mathbb{R}^J$, then $V$ must intersect $U_n$ at some point $x$, which is true if and only if $\pi_\alpha(V)$ intersects $\pi_\alpha(U_n)$ at $\pi_\alpha(x)$ for every $\alpha \in J$. Hence, $\pi_{\alpha_i}(U_n)$ is a dense open subset of $\mathbb{R}$ for every $\alpha_i \in F$.
By exercise $6(a)$ of section 43, a countable product of completely metrizable spaces is completely metrizable. Thus, $\mathbb{R}^F$ is completely metrizable and hence Baire. Further, because any finite product of open dense sets is itself dense, $\bigcap_n \pi_F(U_n)$ is dense in $\mathbb{R}^F$, where
$$\pi_F(U_n) = \prod_{\alpha_i \in F} \pi_{\alpha_i}(U_n).$$
More specifically, there is at least one point $x \in \mathbb{R}^F$ such that $\pi_{\alpha_i}(x) \in \pi_{\alpha_i}(V) \cap \pi_{\alpha_i}(U_n)$ for every $n$ and every $\alpha_i \in F$.
It remains only to show that, for $\alpha \in J - F$, there is some $y_\alpha \in \bigcap_n \pi_\alpha(U_n)$ (because $\pi_\alpha(V) = \mathbb{R}$). Then, letting $z$ be such that $\pi_{\alpha_i}(z) = \pi_{\alpha_i}(x)$ and $\pi_{\alpha}(z) = y_\alpha$, we find that $z \in V \cap \bigcap_n U_n$. But how do I know that $\bigcap_n \pi_\alpha(U_n)$ is nonempty for any $\alpha \in J - F$? This answer suggests I can assume $U_n \supset U_{n+1}$ "without loss of generality", but I don't see how that is so.
I greatly appreciate any advice. I've been juggling products, projections, and intersections in my brain for so long I'm starting to forget my own name!
 A: One cannot prove the product of Baire spaces is Baire; this might even fail for two spaces. (Oxtoby found/constructed some examples of this). But $\mathbb{R}$ is so nice that we can say more:
This paper by Frolík defines a notion of countably complete (def 2.3) : a space $X$ is called countably complete if there is a sequence of bases $\mathcal{B}_n$ for $X$ such that for every nested (i.e. decreasing) family of sets $A_{n_k}$ for some increasing sequence $n_k$ of integers and such that $A_{n_k} \in \mathcal{B}_{n_k}$, the intersection $\bigcap_{k} \overline{A_{n_k}}$ is non-empty. 
Note that all such spaces are Baire: if $U_n$ are open and dense, we can make them decreasing open and dense by defining $V_n = \cap_{i=1}^n U_i$ and noting that all $V_n$ are also open and dense, $V_{k+1} \subseteq V_k$ for all $k$ and $D:= \bigcap_n V_n = \bigcap_n U_n$; so we continue with the $V_n$ instead. 
Now let $O$ be empty and non-empty
Pick $B_n \in \mathcal{B}_n$ with $\overline{B_n} \subseteq B_{n-1} \cap V_n \cap O$ (work by recursion, with $B_0 = X$ as a start), which can be done as we have bases, and $V_n$ is open and dense. The $B_n$ are nested so the intersection of their closures is non-empty, and this lies in $O$ 
and in $\bigcap_n V_n = D$, so $D$ intersects every non-empty open set of $X$, hence is dense. So $X$ is Baire.
It's also clear that $\mathbb{R}$ has this property. Take $\mathcal{B}_n$ to be all 
open intervals of diameter $\le \frac{1}{n}$, and Cantor's theorem for complete 
metric spaces shows that $\mathbb{R}$ is countably complete for this choice.
Theorems 2.10 and 2.12 state that any (box) product of countably complete spaces is countably complete. The proof basically is: take all base sets formed from members of the $n$-th bases in the component spaces, and use this $n$-th base for the product.
So $\mathbb{R}^J$ is Baire in the product and the box topology, and in the uniform topology it's even completely metrisable, hence Baire. 
