For fixed $g$ note that $\sup\limits_{x,y\in T} \langle g,x-y \rangle = \sup\limits_{x,y\in T} |\langle g,x-y \rangle|$, hence
$$\left(\sup\limits_{x,y\in T} \langle g,x-y \rangle \right)^2=\left(\sup\limits_{x,y\in T} |\langle g,x-y \rangle|\right)^2=\sup\limits_{x,y\in T} |\langle g,x-y \rangle|^2=\sup\limits_{x,y\in T} \langle g,x-y \rangle^2$$
Let $F:g\mapsto \sup\limits_{x,y\in T} \langle g,x-y \rangle$. The previous equalities show that $h(T-T)^2=\mathbb E(F(g)^2)$, and of course $w(T-T)=\mathbb E(F(g))$.
Let us prove that $F$ is $\mathrm{diam}(T)$-Lipschitz: for $g,g'\in \mathbb R^n$,
$$\langle g,x-y \rangle = \langle g-g',x-y \rangle + \langle g',x-y \rangle \leq \|g-g'\|\mathrm{diam}(T) + F(g')$$
hence $F(g) - F(g')\leq \|g-g'\|\mathrm{diam}(T)$ and the claim is obtained by symmetry.
Gaussian concentration provides an upper bound on $\mathbb V(F(g))$. Indeed $$\mathbb V(F(g)) = \int_0^\infty P(| F(g)- \mathbb E(F(g))|\geq \sqrt t)\leq 2\int_0^\infty e^{-t/(2 \mathrm{diam}(T)^2)} = 4\mathrm{diam}(T)^2$$
Thus $h(T-T)=\sqrt{\mathbb E(F(g)^2)}\leq \sqrt{w(T-T)^2 + 4\mathrm{diam}(T)^2}\leq w(T-T) + 2\mathrm{diam}(T)$.
Using the Gaussian Poincaré inequality one can get the stronger inequality
$$h(T-T)\leq w(T-T) + \mathrm{diam}(T)$$
Regarding the other inequalities, $w(T-T)\leq h(T-T)$ follows from Jensen's inequality:
$$h(T-T)=\sqrt{\mathbb E\left[\left(\sup\limits_{x,y\in T} |\langle g,x-y \rangle| \right)^2\right]}\geq \mathbb E (\sup\limits_{x,y\in T} |\langle g,x-y \rangle|) = w(T-T)$$
The last inequality $w(T-T)+2\mathrm{diam}(T) \leq Cw(T-T)$ follows from Proposition 7.5.2 of the book:
$$w(T-T)+2\mathrm{diam}(T)\leq w(T-T)+ 2\sqrt{2\pi}w(T) = \left(1+\sqrt{2\pi} \right)w(T-T)$$
Using the tighter bound on the variance, the last constant can be improved to $1+\sqrt{\frac \pi 2}$.
For (a), one approach is to show that the distribution of $\|Px\|_2$ where $P$ is random and $x$ is fixed is the same as the distribution of $\|Px\|_2$ when $P$ is fixed and $x$ is random. All this really uses is rotational invariance (i.e., invariance under orthonormal changes of basis).
Formally, we proceed as follows: Let $S\in O(n)$ be fixed, and let $P$ be drawn randomly from the Grassmanian manifold $G_{n,m}$. Because multiplication by $S$ is an orthonormal change of basis and the Grassmanian does not depend on choice of basis, $P\sim S^\top P S$. Because this is true for any fixed $S$, if we take $T\sim \mathrm{Unif}(O(n))$, then $P\sim T^\top P T$ as well. So
$$ \|Px\|_2\sim \|T^\top PT x\|_2 = (PTx)^\top TT^\top (P T x) = \|PTx\|_2. $$
Let $y = Tx$. I claim $\|Py\|_2\sim\|\pi y\|_2$ for any fixed $P\in G_{n,m}$, where $\pi\in G_{n,m}$ is the projection onto the first $m$ coordinates. Since $P$ is a projection, it has singular value decomposition $P = S^\top\pi S$ for $S\in O(n)$. It follows that $\|Py\|_2=\|\pi S y\|_2\sim\|\pi y\|_2$, where we use the fact that $S T \sim T$ because $\mathrm{Unif}(O(n))$ is invariant under orthonormal changes of basis. To finish, observe that we can write $Q = \pi T$, so $Qx\sim \pi Tx = \pi y$.
For (b), we have $Q^\top z\sim (QT)^\top z = T^\top (Q^\top z)$, for $T\sim\mathrm{Unif}(O(n))$, using similar reasoning as above.
Best Answer
Check the latest version of the book (it's just on his website), he corrected this problem in the latest version. Now you are asked to "Show that the subgaussian norm of this distribution is not bounded by an absolute constant as the dimension n grows." There’s nothing wrong with this argument.
But what worth noticing is that you need to have a finer bound in the case where your bounds on $\langle X,x \rangle$ depends on $n$. Because the orcliz norm may not grow as the dimension $n$ and the size of convex body grows. Like the case when $X\sim \text{Unif}(\sqrt{n} S^{n-1})$, you need a better bound for it. But if you are only interested in Unif$(S^{n-1})$ (say), using the boundedness should be just fine.