Let $B \subset X$ where $X$ is a normed linear space over $\mathbb{R}$ and $B$ is a proper subspace. If $B$ is not closed, is $B$ necessarily a hyperspace(maximal proper subspace) in $X$. I attempted to prove this.
Attempted proof:
Choose a basis $\mathfrak{B}$ of $B$ and extend it to a basis $\mathfrak{X}$ of $X$.
Let $f \colon \mathfrak{X} \to \mathbb{R}$ be the map which is $0$ on $\mathfrak{B}$ and $1$ on $\mathfrak{X} \setminus \mathfrak{B}$
$ f(x) = \begin{cases} \text{0} &\quad\text{if x $\in \mathfrak{B}$}\\ \text{1} &\quad\text{otherwise}\\ \end{cases}$
This map being defined on basis extend linearly to the whole space.
By 1-1 correspondence of submodules, the kernel of above map is a hyperspace. Hence $B$ is a hyperspace.
What is wrong here?