If $B$ is a subset of a Banach space is it true that if $\operatorname{span} B$ is closed then $\operatorname{span} B$ is finite dimensional Let $V$ be a separable Banach space
Let $B \subset V$ be a linearly independent, closed and bounded subset of $V$
I would like to know if is it true that
$$
\operatorname{span} B
\text{ is closed }
\Longrightarrow 
\operatorname{span} B
\text
{ is finite dimensional}
$$
thanks
 A: Yes, the claim is true.
There are some nice results in the paper

Bartoszyński, Tomek; Džamonja, Mirna; Halbeisen, Lorenz; Murtinová, Eva; Plichko, Anatolij, On bases in Banach spaces, Stud. Math. 170, No. 2, 147-171 (2005). ZBL1093.46012.

Let's let $W = \operatorname{span} B$; note that $W$ is also separable.  Suppose $B$ is linearly independent and $W$ is closed and infinite dimensional; then $B$ is a Hamel basis for $W$.  Note that $B$ is closed in $W$ iff it is closed in $V$.  It is shown in Theorem 3.10 of the above paper that a Hamel basis for an infinite-dimensional separable Banach cannot be an analytic set, and in particular it cannot be closed.
(There should be a much more elementary argument to just show directly that it can't be closed.  I will think about it.)
If you don't assume $V$ is separable then it can fail.  Theorem 3.8 of the above paper shows that for sufficiently large Hilbert spaces ($V = \ell^2(\mathbb{R})$ would do), it is possible to construct a Hamel basis which is closed.  
