Find on $C[0,1]$ closed and bounded set $A$ that there are no such $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$

Question

Find in $C[0,1]$ closed and bounded set $A$ such that there are no $f,g \in A$ that imply $\operatorname{diam}(A)=d(f,g)$, where $\operatorname{diam}(A) = \sup\{d(f,g)\mid f,g \in A\}$.

Answer 1

Define $f_n$ as:

$f_n = 1-(1/n)$ on $[1/n, 1]$, $0$ in $0$, and linear on the rest (but so that the function is continuous, basically connecting the points $(0,0)$ and $(1/n,1-(1/n))$ on the graph of function f_n).

Now when considering the set $A=\{f_n: n \in \mathbb{N} \}$:

$f_1$ will be costant $0$ function, and the other functions will be going up in their maximum, which will be $1-(1/n)$ - so $diam(A)=1$, but no two functions have that distance between them.

The set is bounded, and any sequence in it that isn't eventually constant, does not converge to anything in the the space of continuous functions with $sup$ metric, as the pointwise limit of such a sequence would be a function $f$ with values $0$ in $0$, and $1$ on $(0,1]$, and that is not a continuous function.