Show that every (separable) infinite-dimensional Banach space $X$ contains a proper vector subspace $Y$ with $cl (Y)$ = $X$
This is Exercise 1 (f) p.245 of this book.
Assuming $X$ is separable, there is an increasing sequence of finite-dimensional subspaces $G_n$ such that $G:=\bigcup_n G_n$ is dense in $E$. The vector subspace $span(G)$ is dense as well.
Since $X$ has infinite dimension, each $G_n$ is closed and has empty interior. By Baire's theorem, $G$ has empty interior. However this does not imply that $span(G)\neq X$.
Can this line of thought lead to a solution ? Completely different solutions are fine.
I'm also interested in a proof for the case where $X$ is not separable.
Best Answer
Your spaces $G_n$ are non decreasing so $G$ is already a dense vector subspace. You may then apply Baire's category theorem to conclude that $G$ has empty interior.
In general when $X$ is an infinite dimensional real normed vector space it has a dense hyperplane. You can show it in three steps:
For all $\varphi : X \longrightarrow \mathbb R$ linear, either $N(\varphi) = \operatorname{Ker} \varphi $ is closed; either it is dense.
For all $\varphi : X \longrightarrow \mathbb R$ linear, $\varphi$ is continuous if and only if its kernel is closed.
There exists a linear non continuous form on $X$.