Was reading Munkres' Topology and got stuck in this. He proves that uniform topology is contained in box topology on $\mathbb{R}^J$ but leaves the case that this inclusion is strict when $J$ is infinite, as an exercise. I could prove the case when $J=\mathbb{N}$ as follows:
If $J=\mathbb{N}$ then consider the basis element $\prod\limits_{i\in \mathbb{N}}(0,1)$ of the box topology. Now let $\vec{x}:=(1/2,2/3,3/4,...,n/(n+1),...)$(or any sequence in $(0,1)$ increasing to $1$ will do!). Clearly $\vec{x}\in \prod\limits_{i\in \mathbb{N}}(0,1)$. We show that $\prod\limits_{i\in \mathbb{N}}(0,1)$ is not open in uniform topology. For if it were then there would be an $\epsilon$-ball ( with $\epsilon <1$ without loss) (under the metric $\bar{\rho}(\vec{x},\vec{y}):=\sup\limits_{ i\in \mathbb{N}} (\min (1,|x_i-y_i|))$) around $\vec{x}$ s.t $B_{\bar{\rho}}(\vec{x},\epsilon)\subset \prod\limits_{i\in \mathbb{N}}(0,1)$. But we can choose $y_N$ such that for large enough $N$, $x_N$ is very close to $1$ and $y_N>1$ and $|x_N-y_N|<\epsilon /2$. Now define $\vec{y}$ as $y_i=x_i$ iff $i\neq N$. Then $\bar{\rho}(\vec{x},\vec{y})=|x_N-y_N|<\epsilon /2<\epsilon$ so $\vec{y}\in B_{\bar{\rho}}(\vec{x},\epsilon)$ but clearly $\vec{y}\notin \prod\limits_{i\in \mathbb{N}}(0,1)$. So $\prod\limits_{i\in \mathbb{N}}(0,1)$ is not open in uniform topology on $\mathbb{R}^\mathbb{N}$.
My question is that is there a way to generalize this proof for the general case where $J$ can be uncountable?? Also if one thinks that my proof can be improved feel free to comment. Thanks in advance.
$(0, 1)^{\mathbb{N}} \times \mathbb{R}^{J\backslash{\mathbb{N}}}$ will be open in the box topology but not in the uniform metric topology if $J$ is an infinite set that contains $\mathbb{N}$. The argument is basically the same as the one you gave.