I'm trying to prove below result which is mentioned in page 24 of Santambrogio's Optimal Transport for Applied Mathematicians: Calculus of Variations, PDEs, and Modeling. Could you have a check on my attempt?
Theorem: Let $E$ be a normed space and $f:E \to \mathbb R$ strictly convex. Then $\partial f (a) \cap \partial f (b) = \emptyset$ for all $a\neq b$.
Proof: We need the following useful result whose proof is given at the end.
Lemma: If $\varphi \in \partial f (a)$ then $f(x) - f(a) > \langle \varphi, x-a \rangle$ for all $x\neq a$.
Assume the contrary that there are $a \neq b$ such that there is $\varphi \in \partial f (a) \cap \partial f (b)$. Then by our Lemma, $$ f(b) - f(a) > \langle \varphi, b-a \rangle \quad \text{and} \quad f(a) - f(b) > \langle \varphi, a-b \rangle. $$
It follows that $$ f(b)-f(a) > \langle \varphi, b-a \rangle> f(b)-f(a), $$
which is a contradiction. This completes the proof.
Proof of the lemma: Assume the contrary that there is $b \neq a$ such that $f(b) - f(a) = \langle \varphi, b-a \rangle$. Let $c := (a+b)/2$. Then $$ \frac{f(a)+f(b)}{2} > f(c) \ge f(a) + \langle \varphi, c-a \rangle. $$
It follows that $$ \frac{2f(a) + \langle \varphi, b-a \rangle}{2} > f(a) + \langle \varphi, c-a \rangle. $$
So $\langle \varphi, b-a \rangle > \langle \varphi, 2(c-a) \rangle$. On the other hand, $2(c-a) = b-a$. Then we obtain a contradiction.