The original question I was given was
Let $\mathbf{H}$ be an infinite Hilbert space. Is it possible that a countable subset of $\mathbf{H}$ spans it?
Assuming $\mathbf{H} = \mathrm{Sp}B$ where $B$ is a countable set, $\mathbf{H}$ is separable and therefore isomorphic to $\ell_2$, and $\left \{ Au\mid u\in B \right \}$ (where $A$ is the isomorphism between $\mathbf{H}$ and $\ell_2$) is a countable subset of $\ell_2$ that spans it. I want to say that this is not possible, which will make it impossible in the general case.
If I knew this subset is also linearly independent, I could use Gram-Schmidt to obtain an orthonormal basis $\left ( \varphi_k \right)_{k=0}^\infty$ such that $\mathrm{Sp}\left \{ \varphi_k\right \} = \ell_2$, and then the vector $\sum\limits_{k=0}^\infty \frac{1}{k^2}\varphi_k$ would be in $\ell_2$ but not in $\mathrm{Sp}\left \{ \varphi_k\right \}$. However I can't find a way to do it without the set being linearly independent. Does it necessarily have a linearly independent subset? I know this is a thing in finite dimensional spaces, but is it true also in (countably) infinite dimensions?
In any vector space any set of vectors $B$ contains a maximal linearly independent subset $C$ (by Zorn's Lemma). It then follows by maximality that $B$ and $C$ span the same subspace of $V$. Of course the isomorphism $A$ preserves linear independence. Incidentally, no Banach space can be spanned by countable number of elements. (This is proved using Baire Category Theorem).