I have written the following proof of the fact that,
$\Bbb R^J$ is Baire space in product topology.
Can anyone check my proof and say about any fault? Thanks in advance.
$\textbf{Proof :---}$ Here we only consider product topology. So let $\{U_i\}$ be a countable collection of open dense subsets of $\Bbb R^J$. We show the intersection $\bigcap_{i\in \Bbb N}U_i$ is dense in $\Bbb R^J$. Define $W_n=\bigcap_{i=1}^nU_i$ for each $n\in\Bbb N$. Then $W_n$ are open dense in $\Bbb R^J$ and $\bigcap_{i\in \Bbb N}U_i=\bigcap_{n\in \Bbb N}W_n$.
Choose any non-empty $V\subseteq_{\text{open}}\Bbb R^J$. Since $W_1$ is open dense we have $W_1\cap V$ is also open, so $W_1\cap V$ contains some non-empty basic open sets of the form $\bigcap_{j\in F_1}\pi_j^{-1}(c_{j,1},d_{j,1})$, where $c_{j,1}$ and $d_{j,1}$ are reals and $F_1$ is a finite subset of $J$ and $\pi_j$ is the projection on the $j$-th coordinate. Now any non-empty open interval on $\Bbb R$ contains a closed interval. So for each $j\in F_1$, we have real numbers $a_{j,1}$ and $b_{j,1}$ for $j\in F_1$ such that $\bigcap_{j\in F_1}\pi_j^{-1}[a_{j,1},b_{j,1}]\subseteq\bigcap_{j\in F_1}\pi_j^{-1}(c_{j,1},d_{j,1})\subseteq W_1\cap V$.
Now $W_2$ is open dense. So $W_2\cap\bigg(\bigcap_{j\in F_1}\pi_j^{-1}(a_{j,1},b_{j,1})\bigg)$ is non-empty open set. So $W_2\cap\bigg(\bigcap_{j\in F_1}\pi_j^{-1}(a_{j,1},b_{j,1})\bigg)$ contains some non-empty basic open sets of the form $\bigcap_{j\in F_2}\pi_j^{-1}(c_{j,2},d_{j,2})$, where $c_{j,2}$ and $d_{j,2}$ are reals and $F_2$ is a finite subset of $J$. Now any non-empty open interval on $\Bbb R$ contains a closed interval. So for each $j\in F_2$, we have real numbers $a_{j,2}$ and $b_{j,2}$ such that $\bigcap_{j\in F_2}\pi_j^{-1}[a_{j,2},b_{j,2}]\subseteq\bigcap_{j\in F_2}\pi_j^{-1}(c_{j,2},d_{j,2})\subseteq W_2\cap\bigg(\bigcap_{j\in F_1}\pi_j^{-1}(a_{j,1},b_{j,1})\bigg)\subseteq W_2\cap\big(W_1\cap V\big)=W_2\cap V$.
Now $W_3$ is open dense. So $W_3\cap\bigg(\bigcap_{j\in F_2}\pi_j^{-1}(a_{j,2},b_{j,2})\bigg)$ is non-empty open set. So $W_3\cap\bigg(\bigcap_{j\in F_2}\pi_j^{-1}(a_{j,2},b_{j,2})\bigg)$ contains some non-empty basic open sets of the form $\bigcap_{j\in F_3}\pi_j^{-1}(c_{j,3},d_{j,3})$, where $c_{j,3}$ and $d_{j,3}$ are reals and $F_3$ is a finite subset of $J$. Now any non-empty open interval on $\Bbb R$ contains a closed interval. So for each $j\in F_3$, we have real numbers $a_{j,3}$ and $b_{j,3}$ such that $\bigcap_{j\in F_3}\pi_j^{-1}[a_{j,3},b_{j,3}]\subseteq\bigcap_{j\in F_3}\pi_j^{-1}(c_{j,3},d_{j,3})\subseteq W_3\cap\bigg(\bigcap_{j\in F_2}\pi_j^{-1}(a_{j,2},b_{j,2})\bigg)\subseteq W_3\cap\big(W_2\cap V\big)=W_3\cap V$.
Continuing we have, $$\bigcap_{j\in F_1}\pi_j^{-1}[a_{j,1},b_{j,1}]\supseteq\bigcap_{j\in F_2}\pi_j^{-1}[a_{j,2},b_{j,2}]\supseteq\bigcap_{j\in F_3}\pi_j^{-1}[a_{j,3},b_{j,3}]\supseteq...$$ Notice that, $F_1\subseteq F_2\subseteq F_3\subseteq...$ Let $F:=\bigcup_{n\in\Bbb N}F_n$. Then $F$ is countable set.
Next for each $j\in F$ we have $n_j\in\Bbb N$ and a decreasing sequnce $[a_{j,n_j},b_{j,n_j}]\supseteq[a_{j,n_j+1},b_{j,n_j+1}]\supseteq[a_{j,n_j+2},b_{j,n_j+2}]\supseteq...$ of compact sets.\ So $\bigcap_{k\in\Bbb N}[a_{j,n_j+k},b_{j,n_j+k}]$ is non-empty, say $x_j\in \bigcap_{k\in\Bbb N}[a_{j,n_j+k},b_{j,n_j+k}]$. This is because of the fact that, in a compact space a collection of closed subsets having finite intersection property has non-empty intersection. Let $ x\in \Bbb R^j$ be defined as $\pi_j( x)=0$ if $j\in J\backslash F$ and $\pi_j( x)=x_j$ if $j\in F$.
Hence $$ x\in\bigcap_{n\in\Bbb N}\bigg(\bigcap_{j\in F_n}\pi_j^{-1}[a_{j,n},b_{j,n}]\bigg)$$$$\implies V\bigcap\big(\cap_{n\in\Bbb N}U_n\big)=V\bigcap\big(\cap_{n\in\Bbb N}W_n\big)=\bigcap_{n\in\Bbb N}\big(W_n\cap V\big)$$$$\supseteq\bigcap_{n\in\Bbb N}\bigg(\bigcap_{j\in F_n}\pi_j^{-1}[a_{j,n},b_{j,n}]\bigg)\not=\emptyset.$$ So we are done.