I have a question about a step in proving $(M^\perp)^\perp=\overline{M}$ where $M$ is a linear subspace of a normed vector space $E$. And $M^\perp=\{f\in E^*|\langle f,x\rangle =0\}$
This is the proof from Brezis book:
Proof: Assuming the other direction, show $\subset$ direction. Suppose there exists $x_0\in (M^\perp)^{\perp}$ such that $x_0\notin \overline{M}$. Then, Hahn-Banach says there exists a hyperplane that strictly separates $\{x_0\}$ and $\overline{M}$. Thus there are some $f\in E^*$ and $\alpha\in\mathbb{R}$ such that $$ \langle f,x\rangle <\alpha<\langle f, x_0\rangle, \forall x\in M$$
Since $M$ is a linear space it follows that $\langle f,x\rangle =0$(WHY??), for all $x$ and $\langle f, x_0\rangle >0$. Therefore $f\in M^\perp$ and consequently $\langle f, x_0\rangle =0$, a contradiction.
Can someone tell me why the place (WHY??) is true? I couldn't find out the line follows from the fact that $M$ is a linear space.
Thanks
Pick $\beta \in \mathbb R$ then $$\beta \langle f, x\rangle=\langle f, \beta x\rangle < \alpha $$
Since this happens for all $\beta \in \mathbb R$, it follows that $\langle f,x\rangle=0$.