An application of Hahn-Banach: There exists non-zero linear forms vanishing on a subspace I have to prove the following:

Let $E$ be a normed vector space, and $F$ a proper subspace of $E$. Prove that there exists a continuous, non-zero linear form $L$ on $E$ which vanishes on $F$.

This seemed easy at first: I'll define $\ell$ to be zero on $F$, then pick a vector $u$ not in $F$ and extend $\ell$ arbitrarily to a non-zero linear form on $F \oplus \mathbb R u$. Then I'll apply Hahn-Banach to that. But what is the sub-additive function which is going to bound $\ell$ to justify applying H.B.? Certainly not the norm on $E$, since in general $|\ell (x + \lambda u)|=|\lambda|\ell(u)$ is not bounded by the norm of $x+\lambda u$.
 A: Edit: a couple years, later, I just found an argument that requires no theorems, so I'll record it here. I'll leave my previous answer below. 
Assume $\overline F\subsetneq E$. Take $x_0\in  E\setminus\overline F $, and define $\ell_0:\mathbb C x+ F \to\mathbb C$ by $\ell_0(cx+y)=c$. Note that $\|cx_0+y\|>0$ for all $c\ne0$ and all $y\in F $, because if it is zero then $cx_0=-y\in F $. Thus
\begin{align*}
\|\ell_0\|
&=\sup\left\{\frac{|c|}{\|cx_0+y\|}:\ c\in\mathbb C\setminus\{0\},\ y\in F \right\}
=\sup\left\{\frac{1}{\|x_0+y/c\|}:\ c\in\mathbb C\setminus\{0\},\ y\in F \right\}\\ \ \\
&=\sup\left\{\frac{1}{\|x_0+y\|}:\ \ y\in F \right\}
=\frac{1}{\inf\{\|x_0+y\|:\ \ y\in F \}}=\frac1{\operatorname{dist}(x_0,F)}.
\end{align*}
So $\ell_0$ is bounded and by Hahn-Banach there exists an extension $\ell: E\to\mathbb C$ with $\|\ell\|=\|\ell_0\|>0$, and $\ell|_F=0$. 

(old answer)
The result you need is the geometric form of Hahn-Banach (and its usual proof uses the Minkowski functional). Concretely,   if $A=\{u\}$ and $B=F$ are subsets of $E$ with $A$ compact and $B$ closed (you can take here the closure of $F$ which cannot be $E$), there exists a bounded functional $f$ on $E$ with $$ f(u)<f(y),\ \ \ y\in F.$$ 
In particular,  $f(F)$ is a proper real subspace of $\mathbb R$, so $f(F)=\{0\}$, and $f(u)\ne0$. 
