Let $m$ be a plane in $\mathbb{R}^{3}$ and let $R_{p}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is the reflection with respect to the plane $p$.
Let $X \in \mathbb{R}^{3}$ be a countable subset of $\mathbb{R}^{3}$. I would like to prove that there exists a plane $m$ so that $R_{m}(X) \cap X = \emptyset$.
It looks as if there exists a tricky approach using the Baire's theorem, but i do not see any possible way how i can rigorously start, apart from stating some simple observations.
Are there any hints that might help?
I don't think we need Baire. Let $P$ be the $x$-$y$ plane. Write the points of $X$ as $\{(x_n,y_n,z_n): n \in \mathbb N\}.$ Let $u=(0,0,1).$ The desired plane will be $P+tu$ for some $t\in \mathbb R.$
Clearly we need to avoid any $t \in \{z_n: n\in \mathbb N\}.$ We also need to avoid the situation of two distinct points in $X,$ neither in $P+tu,$ that are symmetric with respect to $P+tu.$ In the latter case we will have $t=(z_m+z_n)/2$ for some pair $m,n.$ These are the only examples of a $P+tu$ that doesn't work.
It follows that if
$$t\not \in \{z_n: n \in \mathbb N\} \cup \{(z_m+z_n)/2 : m,n\in \mathbb N\},$$
a countable set, then $P+tu$ has the desired property. There are of course uncountably many such $t.$