What was I thinking
There is a neighborhood W=$\prod_i (-r_i,r_i)$ that does not intersect A . Consider a linear transformation $h_i:x_i \to x_i/r_i$ (linear transformation is a homeomorphism in both the product and box topology on $\mathbb{R^N}$). Also a product of continuous functions is continuous so, that the product of homeomorphisms is a homeomorphism. If y=h(x) $\in$ V , V is open in the box topology, then there is a basis neighborhood y $\in$ V′=$\prod_i V′_i \subseteq V$ and h(z) $\in$ V′ iff for all i : $h_i(z_i) \in V′_i$ iff for all i : $z_i \in h^{−1}_i(V′_i)$ . Therefore, for all z $\in \prod_i h^{−1}_i(V′_i)$ which is open in the box topology if all $h_i$ are continuous, h(z)$ \in V′ \subseteq V$, it is also true for the linear transformation, and if was possible separate 0 from h(A) by a continuous function f , then $f \circ h$ would separate 0 and A . We Know that h(A) is closed and does not have any points in W′=$\prod_i(−1,1)$ for if it had some than A would have a point in W . We also know that $\mathbb{R^N}_{uniform}$ is completely separable, and there is a continuous function f that separates 0 and X−B(0,1) $ \subseteq $ W′. So, f is continous in the finer box topology as well, and separates 0 from h(A). This way we show that 0 can be separated by a continuous function from a closed set A that does not contain it.
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If $\langle X_i:i\in I\rangle$ is any $I$-indexed family of regular spaces, then the box product $Y=\prod_{i\in I}X_i$ is regular. To see this, let $y=\langle x_i:i\in I\rangle \in Y$, with $C\subseteq Y\setminus\{y\}$ a closed set. Then it is easy to check that, for each $i\in I$, there is a closed $C_i\subseteq X_i$ with $C\subseteq \prod_{i\in I}C_i\subseteq Y\setminus\{y\}$. For each index $i$, we know that $x_i\not\in C_i$; so, using regularity of $X_i$, let $U_i,V_i$ be disjint open sets in $X_i$ such that $x_i\in U_i$ and $C_i\subseteq V_i$. Then the sets $U=\prod_{i\in I}U_i$ and $V=\prod_{i\in I}V_i$ are open in the box topology, they're disjoint, $y\in U$, and $C\subseteq V$. Thus $Y$ is regular.