The intersection of closure of span of infinite, linearly independent, closed, bounded, separated subsets of $\ell^2$ Let $X$ and $Y$ be two subsets of $\ell^2$ space over $\mathbb{C}$ such that: $X \cup Y$ is linearly independent, $X \cap Y = \emptyset$ and $\inf_{x \in X, y \in Y} \| x-y \|>0$ and such that each of them is: infinite, closed and bounded.
My question is if is it true that
$$
\overline{
\operatorname{span}
X
}
\cap
\overline{
\operatorname{span}
Y
}
=
\{0\}
$$
 A: No. Here's a simple example.
In $\ell_2$ with orthonormal basis $(e_n)_{n\ge 1}$ choose $X=\{e_n:n\ge 1\}$ (so $X$ is closed and bounded). Then choose a sequence $v_n$ of vectors with $\|v_n\|\le 1/2$ and define $f_n=e_{3n-2}+e_{3n-1}+v_n$, $Y=\{f_n:n\ge 1\}$. Then $Y$ is also closed (discrete) and bounded, and $d(f_n,e_m)\ge 1/2$ for all $n,m$. Since $\mathrm{Span}(X)$ is dense, the intersection 0 condition does not hold.
Then it is enough to find $(v_n)$ such that $X\cup Y$ is linearly independent. We choose $v_n=\sum_{k}u_{n,k}e_{3k}$. This already ensures that $(f_n)$ is linearly independent. Write $P_n=\{k:u_{n,k}\neq 0\}$. We require that $P_n$ is infinite and $P_n\cap P_m=\emptyset$ for all $n\neq m$. Then this ensures that $X\cup Y$ is linearly independent: indeed, write a linear relation as $\sum_{i=1}^n\lambda_ie_i=\sum_{i=1}^m\mu_if_i$, with $\mu_m\neq 0$. Then there exists $k\in P_m$ with $k>n$. Projecting orthogonally on $e_k$ then yields $0=\mu_mu_{m,k}$, a contradiction.
