Whether each dense linear subspace of a separable Banach space has finite co-dimension? This question is a sort of converse to this post.  If $E$ is an infinite-dimensional separable Banach space and $F$ is a dense (infinite-dimensional) (linear) subspace, then is $E/F$ of finite co-dimension?  (I know it will not be Hausdorff since $M$ is not closed (unless it is equal to $E$ itself), but this doesn't matter).  
 A: Let $E=\ell_2$, $F=\ell_{2f}$, that is a set of sequences $(x_n)_{n\in\Bbb N}\in\ell^2$ which are eventually zero. Then $F$ is dense in $E$, but a vector space $E/F$ is not finitely-dimensional, because $E\ne A+F$ for each finite dimensional subspace $A$ of $F$. Let us prove the last claim. We have $F=\bigcup F_n$, where $F_n$ consists of all vectors $x=(x_i)\in\ell_2$ such that $x_i=0$ for each $i>n$. Each $F_n$ (and so $F’_n=E_n+A$) is a finitely-dimensional subspace of $\ell_2$. 
We claim that $F’_n$ is nowhere dense in $\ell_2$ for each $n$.  Indeed, for each $\delta>0$ put  $B_\delta=\{x\in\ell_2: \|x\|\le\delta\}$ and suppose to the contrary that $F’_n$ is dense in a ball $x+B_{3\varepsilon}$, where $x\in\ell_2$ and $\varepsilon>0$. Let $\{e_i\}$ be the standard unit vectors of the space $\ell_2$. Then $x+3\varepsilon e_i\in x+ B_{3\varepsilon}$ for each $i$. Then for each $i$ there exists $x_i\in F’_n$ such that $\|(x+3\varepsilon e_i)-x_i\|\le\varepsilon$ (and so $x_i\in  x+B_{4\varepsilon}$). It follows that for distinct $i,j$ we have $$\|x_i-x_j\|\ge \|(x+3\varepsilon e_i)- (x+3\varepsilon e_j)\|- \|(x+3\varepsilon e_i)-x_i\|-\|(x+3\varepsilon e_j)-x_j\|\ge$$ $$3\varepsilon-\varepsilon-\varepsilon=\varepsilon.$$
Thus $\{x_i\}\subset x+B_{4\varepsilon }$ is a bounded but not totally bounded subset of $F’_n$. But this is impossible, because it is well-known that any finitely dimensional topological linear space over $\Bbb R$ is topologically isomorphic to  the space  $\Bbb R^n$ endowed with the standard topology, and so $F_n’$ is locally compact. 
It remains to remark that by Baire theorem a complete metric space $\ell_2$ cannot be a union of a countable family $\{F_n’\}$ of nowhere dense subsets. 
