Closed subspace. A Hahn–Banach theorem consequence

Question

I am trying to prove:

If M is a subspace of a normed space $X$, that $\overline{M}=\bigcap\{\ker(\phi):\phi|_{M} = 0 \}$

It is really easy to see that $\overline{M} \subset \bigcap\{\ker(\phi):\phi|_{M} = 0 \}$. However, I don't know how to use the Hahn-Banach theorem to prove the other inclusion.

Answer 1

For each $x\notin \overline{M}$, as a consequence of Hahn-Banach Theorem, there is $\phi_x \in X^*$, $\|\phi_x \|=1$, $\phi_{x}|_{M}=0$, $\phi_x(x)=d(x,\overline{M})$.

If $y\in \bigcap\{\ker(\phi): \phi|_M=0\} \subset \bigcap\{\ker(\phi_x)\}$, then $\phi_x(y) = 0 \quad \forall x \notin \overline{M}$. Thus $y\in \overline{M}.$