Suppose $X$ is an infinite dimensional topological vector space 
and $v\in X$ is non-zero. It is then not difficult to construct 
a vector space $U\subset X$ so that


1) $U$ is dense in $X$.


2) $U+{\mathbb C} v = X$.


In particular $U$ has co-dimension 1 in $X$ but is not closed.


Now, proofs that I can think of uses Zorn: Consider/construct a
dense subspace $V$ not containing $v$ (usually not so difficult)
and consider the collection of subspaces containing $V$ but not $v$,
partially ordered under inclusion. If a vector $w\in X\setminus (V+{\mathbb C} v)$ you may add it to $V$ to get a larger such space.
It is strictly inductively ordered so a maximal element $U$ will
do the job.


The catch is that this is not very 'visual'. I can not give
an explicit description of any such maximal element.
So my question is if there is an explicit way to construct such a $U$
without resorting to Zorn/Axiom of Choice?


If possible, best would be an example with $X$ a separable Banach/Hilbert space.



Construction of a codimension 1 dense subspace without Zorn
You are asking for a discontinuous linear functional that is non zero at $v$, but it is consistent with $ZF$ that every linear functional on a Banach space is continuous. However, on some normed spaces you can do what you want in $ZF$. For example, take $X:=c\_{00}$, the space of finitely non zero real sequences under the sup norm. Given $0\not= v$ in $X$, you can explicitly construct an element $u$ in $\ell\_2 \sim \ell\_1$ s.t. $\langle u, v\rangle =1$. 


