This even fails for $E=\mathbf R$; take $f(x)=|x|$. And it is true if $df(x_0;v)$ is defined using $\lim_{\varepsilon\to 0}$.

There seems to be a typo in the definition of a $p$-nuclear operator: one should have $\|(y_n)_n\|_q^w<\infty$, where $1/p+1/q=1$. Likewise, the expression for $\nu_p(T)$ should be corrected.

Correction: Not $S$ is weakly compact, but its unit ball...

Now, every $Y$ as above is a quotient of $L_1[0,1]$, and hence every $X$ as above is a subspace of $L_\infty$. Thus, $S$ is a weakly compact subset of $L_\infty$ and therefore norm separable.

Maybe the OP wants to consider the unit ball rather than the unit sphere, which in light of Bill's comments is more reasonable to consider?

If $(|\lambda_n|)$ is decreasing and in $\ell_2$, then $\sum_{n=1}^\infty |\lambda_n|^2 \ge \sum_{n=1}^N |\lambda_n|^2 \ge N |\lambda_N|^2$.

@Bill, probably it's this paper by Wolfgang Ruess that you have in mind: zbmath.org/?q=an%3A0573.46007

My apologies; the part of my previous comment that the $C_n$ must be empty was bare nonsense.

If $\Delta$ is as suggested by Robert, then, again by one of the corollaries of Baire's theorem, $\chi_\Delta$ has a point of continuity (being Baire-$1$), which it doesn't. I think, Robert's $C_n$ are all empty (hence closed!), since $f_n$ is continuous; still the convergence assumption would force the union of the $C_n$ to be $[0,1]$. -- As for notation, it seems to me that the OP has defined his own ``weakish'' convergence different from weak convergence in Banach spaces.

To separate a point $x$ from a closed convex set $C$, one can take the best approximation $y\in C$ to $x$ and a hyperplane perpendicular to $x-y$.

Good point! (I have deleted my erroneous comment.)

The condition is reasonable; e.g., the constant-1 function in $C[0,1]$ is such an $x$ (and it is extreme). Note that the set of all vectors satisfying the condition in the question is closed; so the best one can hope for is that $x$ is in the (norm-) closure of $E_X$.

Couldn't you argue that $C$ is not diametral because it's compact?

But what are the $e_i$??? What if $n=1$ and $e_1$ is some vector of norm 1000?? [I half guess that you want to refer to the unit vectors; if so, please say so.] And which $p$ do you consider?

Are you assuming anything about the $e_i$, like $\|e_i\|_{p,\infty}\le1$? And is your actual question whether $\|x-y\|_{p,\infty}\le 1$ for $x,y$ in the closed convex hull of $C$?