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No, it isn't true. Let $E$ be a non-Borel set in $X$ and consider the set of Dirac measures $E' = \{ \delta_x : x \in E\} \subset S$. Note that for each Dirac measure $\mu$, the total-variation ball of radius 1/2 about $\mu$ contains no other Dirac measure. Hence $E'$ is closed in $T_N$, hence Borel in $T_N$. On the other hand, it is well known that ...
For sake of completeness, I am writing a full answer based on the suggestion of Pietro Majer. The following are equivalent: 1) $A$ is an isometry w.r.to some norm. 2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1. 3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in ... 11 Operator norms are the same thing as injective tensor norms or, equivalently, smallest dual cross norms on$\operatorname{Hom}(V, W) \simeq V^\ast \otimes W$. This means that the dual norm$\Vert \cdot \Vert^\ast$on$V \otimes W^\ast$has to satisfy the following: $$\Vert x \Vert^\ast = \inf_{x = \sum_i y_i \otimes z_i} \sum_i \Vert y_i \Vert \Vert z_i ... 6 Here is a non trivial constraint : {\rm Hom}(V,W) contains (is spanned by) rank one morphisms$$v\mapsto\ell(v)w,\qquad\ell\in V',w\in W.$$If a given norm over {\rm Hom}(V,W) is induced, then$$\|w\otimes\ell\|=\|w\|_W\|\ell\|_*.$$This yields the necessary condition ... 6 To get dimension 4, I think the norm $$\|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty,$$ where$a=(a_1,a_2,a_3,a_4)$, works. Again$X$samples the unit vector basis uniformly. EDIT: It looks like a variation takes care of dimension 3. Use again $$\|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty,$$... 2 fedja: This is a very nice and elegant answer, thank you! (Have you seen/used such a norm ever before?) Before accepting your answer, I'd like to wait a bit, so that other participants feel more encouraged to present other answers. In fact, let me give here a modification of your answer. I noticed that for$N=3$your norm of$v=(v_1,\dots,v_N)\$ is ...