The intersection of a sequence of bases of a Banach space Let $V$ be a Banach space
Let $\{B_n\}_{n \in \mathbb{N}}$ be a sequence of subsets of $V$ such that:
$$
B_{n+1}
\subsetneq
B_n
$$
and
$$
\bigcap_{n=1}^\infty B_n = \{b_0\}
$$
with $b_0 \in V$ and $\forall n \in \mathbb{N}: B_n$ is linearly independent.
I would like to know if 
$$
\bigcap_{n=1}^\infty 
\overline{\operatorname{span}{B_n}}
=
\operatorname{span}\{b_0\}
$$
Thanks.

The answer is no, in general.
Let $V$ be any Banach space with $\dim V \ge 2$.
Let $x \in V$ be a nonzero vector and let $(x_n)_{n=1}^\infty$ be an injective sequence in $V$ which converges to $x$, such as $x_n = \frac{n}{n+1}x$ for $n \in \mathbb{N}$.
Also, let $y \in V$ be a vector such that $\{x, y\}$ is linearly independent.
Define $B_n = \{x_{n}, x_{n+1}, x_{n+2}, \ldots\} \cup \{y\}$ for $n \in \mathbb{N}$.
We have $B_{n+1} \subsetneq B_n$ since $(x_n)_{n=1}^\infty$ is injective and $\displaystyle\bigcap_{n=1}^\infty B_n = \{y\}$.
However, since $x_n \xrightarrow{n\to\infty} x$, we have $x \in \overline{B_n} \subseteq \overline{\operatorname{span}}B_n$ for every $n \in \mathbb{N}$.
Therefore
$$\{x, y\} \subseteq \bigcap_{n=1}^\infty \overline{\operatorname{span}}B_n $$
but $x \notin \overline{\operatorname{span}}\{y\}$ so $\displaystyle \bigcap_{n=1}^\infty \overline{\operatorname{span}}B_n \ne \overline{\operatorname{span}}\{y\}$.

If $\dim V = 1$ then the statement also isn't true. Let $e \ne 0$ and define $B_n = \{ne, (n+1)e, (n+2)e, \ldots\} \cup \{0\}$.
We have $\displaystyle \bigcap_{n=1}^\infty = \{0\}$, but $$\bigcap_{n=1}^\infty \overline{\operatorname{span}}B_n =  \bigcap_{n=1}^\infty V = V \ne \overline{\operatorname{span}}\{0\} = \{0\}$$
