$\DeclareMathOperator{\diam}{diam}\DeclareMathOperator{\co}{co}\DeclareMathOperator{\Aff}{Aff}$ Let be $E$ a normed vector space (non-Banach). $S \subset E$ bounded.
Remember that $\diam(S) = \sup \{\|y-x\| : x,y \in S \} $. Show that $\diam (co(S)) = \diam S$.
What do you think about this :
Since $S \subset \co(s)$, $\diam (\co(S)) \geq \diam S$.
Let be two convex combinations :
$$ x = \sum\limits_{i=1}^{n} \lambda_i x_i , \ \ \ \ y = \sum\limits_{i=1}^{m} \mu_i y_i $$
Now consider $$V = \co( \{ x_1, ..., x_n \}) \subset \Aff(\{ x_1, ..., x_n \})$$ $$W = \co( \{ y_1, ..., y_m\}) \subset \Aff(\{ y_1, ..., y_m \})$$
$V$ is convex compact in $\Aff(\{ x_1, ..., x_n \})$
$W$ is convex compact in $\Aff(\{ y_1, ..., y_m \})$
$V - W$ is convex compact in $\Aff(\{ x_1, ..., x_n \}) - \Aff(\{ y_1, ..., y_m \})$
So it stems from $\|\cdot \|$ is convex that $\max$ is obtained on an extremal point of $V-W$. We can easily show that an extremal point of $V-W$ is an extremal point of $V$ minus an extremal point of $W$.
Now let's consider a maximiser sequence of $\diam(\co(S))$:
$(u_n)_n, (v_n)_n$ that respects : $\| u_n - v_n\| \to_n \diam(\co(S))$.
We deduce of previously that there exists $(u^{'}_n)_n , (v^{'}_n)_n$ sequences in $S$ such as :
$$\| u_n - v_n\| \leq \| u^{'}_n - v^{'}_n\|$$
Thus : $$ \diam (\co(S)) \leq \diam S $$