TITLE: Hahn-Banach proof by extension of basis
QUESTION [6 upvotes]: Hahn Banach Theorem states that given a linear continuous functional $f$ on a subspace $N$ of a normed space $M$, it can be extended to a linear functional $F$ on the whole space $M$ and the norm of the extension is the same as the one of $f$.
Can someone tell me if something is wrong in the following lines: 
Fact 1: Every vector space has a basis (use axiom of choice if it is infinite dimensional). Fact 2: A linear map is defined if we give the image of a basis.
Take $\{e_i\}_{i\in I}$ a basis of $M$ extending one of $N$, and define $F(e_i):=f(e_i)$ if $e_i$ is in $N$ and $0$ otherwise. This is clearly an extension of $f$, plus the norm of $F$ is the same as that of $f$ because sup $|F(x)|/||x||$ is $0$ for $x \in M \setminus N$ and $||f||$ if $x \in N$. 
In case is correct this seems like a very simple proof of this theorem!

REPLY [3 votes]: Take the space $\mathbb R^2$ with the usual norm.  Choose a non-orthogonal basis.  Apply your method to extend from $1$ dimension to $2$.  See that the norm is not preserved.