Prove that a real Banach Space $X$ is reflexive if and only if each pair of disjoint closed, convex subsets of $X$, one of which is bounded, can be strictly separated by a hyperplane.
The theorem is stated and proved in the book Geometric Functional Analysis and its Applications by Richard Holmes on page-161, which goes like this
We shall show that the disjoint closed convex sets $H$ and $U(M)$ cannot be strictly separated. Suppose otherwise; then there would exist $\psi\in X^*$ and a positive number $\gamma$ such that $\psi(x)<\gamma<\psi(y)$ for all $x\in U(M)$ and $y\in H$. Let us assume that $\phi$ has been extended to all of $X$ via the Hahn-Banach Theorem, and let us call the extension $\phi$ also. Choose any point $\overline{x}\in X$ such that $\phi(\overline{x})=\|\phi\|$ and any $z\in X$ for which $\phi(z)=0$. Then for all $\lambda\in \mathbb{R}, \psi(\overline{x}+\lambda z)>\gamma$. Hence, $\psi(z)=0$ and so $\ker(\phi)\subset \ker(\psi)$. This means that the set $\{\phi,\psi\}$ is linearly dependent. We can therefore choose a constant $\alpha$ so that $\psi=\left(\frac{\alpha \gamma}{\|\phi\|}\right) \phi$. Now if $y\in H$ then $\gamma<\psi(y)=\alpha \gamma$, so that $1<\alpha$. On the other hand, if $x\in U(M)$ then $\psi(x)<\gamma$, and so $\phi(x)=\left(\frac{\|\phi\|}{\alpha\gamma}\right)\psi(x)<\frac{\|\phi\|}{\alpha}$; since $\alpha>1$, this contradicts the definition $\|\phi\|=\sup\{\phi(x):x\in U(M)\}$. I am trying to decipher this proof. It uses the contrapositive logic and is based on the James Theorem a Banach space $X$ is reflexive if and only if every continuous linear functional on $X$ attains its supremum on the closed unit ball in $X$.
And, here is what I could understand in this proof
Let $X$ be a Banach Space and $M$ be a subspace. Let $\phi\in M^*$ be a continuous linear functional. Let $H=\{x\in M:\phi(x)=\|\phi\|\}$ be a hyperplane in $M$ and $U(M)=\{x\in M:\|x\|=1\}$ be a unit ball in $M$. Suppose that there exists a continuous linear functional $\psi\in X^*$ and a positive number $\gamma$ such that $\psi(x)<\gamma<\psi(y)$ for all $x\in U(M)$ and $y\in H.$
Choose a point $\overline{x}\in X$ such that $\phi(\overline{x})=\|\phi\|$, i.e. $\overline{x}\in H$ and any $z\in X$ such that $\phi(z)=0$. Then, for a real number $\lambda$, $\phi(\overline{x}+\lambda z)=\phi(\overline{x})+\lambda \phi(z)$. Since $\overline{x}\in H$ and $\phi(z)=0$, $\phi(\overline{x}+\lambda z)=\|\phi\|+\lambda. 0$. $\implies \phi(\overline{x}+\lambda z)=\|\phi\|$. $\implies \overline{x}+\lambda z\in H$. $\implies \psi(\overline{x}+\lambda z)>\gamma$. What I couldn't understand is
How the idea of extending $\phi$ is applied in all of this?
How do you infer $\psi(z)=0$?