Limit point of basic sequence in Banach space This question is about Schauder bases in Banach spaces.
Suppose $(x_n)$ is a basic sequence in a Banach space $E$. Prove that $0$ is the only possible weak limit point of $\{x_n \in E \;\vert\; n \in \mathbb{N}\}$.
 A: Edit: I was initially assuming that $E$ is the closed linear span of the basic sequence, but then I realized that from your question there is no justification for me having done so.  Since my initial argument applied only to that case and I didn't know how to extend it, I did a Google search to find that one way to reduce to this case is to apply Mazur's theorem, and that a nice and succinct answer to your question is given as Lemma 1.6.1 in Topics in Banach space theory by Albiac and Kalton.  (This is something I should have known, but I had apparently not thought enough about weak topologies.)

For what it's worth, here's my original argument that applies after reducing to the case where $E$ is the closed linear span of $\{x_n\}$.
Thus every element $x$ of $E$ has a unique representation as an infinite sum $x=\sum_n a_n(x)x_n$, and the coefficient functionals $x\mapsto a_n(x)$ are bounded.  Let $B=\{x_n\}$ denote the basis, and suppose that $x$ is in $\overline{B}\setminus B$.  You want to show that $x=0$, which is equivalent to showing that $a_n(x)=0$ for all $n$. 
For each $n$ and each $\varepsilon\gt0$, consider the subbasic neigborhood $U_{n,\varepsilon}$ of $x$ defined by  $U_{n,\varepsilon}= \{y:|a_n(y)-a_n(x)|\lt \epsilon \}$.  Suppose to get a contradiction that $x\neq 0$, so there exists a $k$ such that $a_k(x)\neq 0$.  Then $U_{k,\varepsilon}$ is disjoint from $B\setminus\{x_k\}$ for sufficiently small $\epsilon$, and since $x$ is in $\overline B$ you can deduce that $a_k(x)=1$.  Since $x\neq x_k$, there exists $m\neq k$ such that $a_m(x)\neq0$, and the same argument indicated above shows that $a_m(x)$ must be $1$.  Then $U_{k,1}\cap U_{m,1}$ is a neighborhood of $x$ that doesn't intersect $B$, a contradiction.
