Construct dense subspace of codimension $n$ for all $n$ I want to prove the following:
Let $X$ be an infinite dimensional normed space. For all integer $n\geq1$: $X$ has a dense subspace of codimension $n$, i.e. a subspace $L$ such that $\dim(X/L)=n$.
How can i do this? My first thought was: Take a basis $e_1,e_2,\ldots$ such that $X=Span(e_1,e_2,\ldots)$ and let $L=Span(e_2,e_3,\ldots)$ than $\dim(X/L)=1$. On the same way you can go on. Is this the right method to conclude the result?
Thank you.

You can prove this by induction.
Basis of induction: Since $X$ is infinite dimensional there exists a discontinuous linear functional $f$ on $X$. Since $f$ is discontinuous, then $\ker f$ is dense in $X$. As $f$ is a functional we have $\dim (X/\ker f)=1$. 
Step of induction: Assume we have constructed a dense subspace $X_n$ of codimension $n$, then apply the argument given above to get a dense in $X_n$ subspace $E$ of codimension $1$ in $X_n$. Obviously, $E$ is of codimension $n+1$ in $X$ and still dense in $X$. Now set $X_{n+1}:=E$.
