Consider a linearly independent sequence $(x_n)$ in a separable Banach space. Can we always find a closed hyperplane $H$ and and a finite non empty subset $F$ of $\mathbb{N}$ such that $\{x_j: j\in \mathbb{N}\setminus F\}\subset H$. In others words, $H$ contains all but possibly finitely many members of the sequence.
If the closed span of the sequence is not the entire Banach space, then the answer is clearly yes.
My question seems to be connected to this one:
Let $X$ be a Banach space with a normalized basis $(e_n)_{n=0}^{\infty}$ and consider, for each $n$, the vectors:
$$ x_n:=e_0+\frac{\lambda_n}{1!}e_1+\frac{\lambda_n^2}{2!}e_2+\dots... $$
Where $\lambda_n\to 0$, and $\lambda_p\neq\lambda_n\neq 0$ for any $n,p$. It is easy to see that $(x_n)$ is linearly independent.
Let $(x_{n_k})$ be any subsequence of $(x_n)$. Then:
$$ x_{n_k}\to e_0, \text{ so } e_0\in[x_{n_k}] $$
$$ \frac{1!}{\lambda_{n_k}}(x_{n_k}-e_0)\to e_1, \text{ so } e_1\in[x_{n_k}] $$ and so on, we obtain that for any $n$, $e_n\in[x_{n_k}]$. Therefore $[x_{n_k}]=X$. This means that any hyperplane can contain at most finitely many vectors form the sequence $(x_n)$.