I'm currently reading through Roman Vershynin's High Dimensional Probability and working through one of the exercises (7.6.1). Consider a set $T \subseteq \mathbf{R}^n$ and define its Gaussian width $w(T)$, as
$$ w(T) := \mathbb{E} \sup_{x \in T} \langle g, x\rangle, \quad g \sim \mathcal{N}(0, I_n). $$
A closely related version, $h(T)$, is defined similarly:
$$ h(T) := \sqrt{\mathbb{E}\left[ \sup_{x \in T} \langle g, x \rangle^2 \right]}. $$
Now, Exercise 7.6.1 in the book asks the reader to show that
$$ h(T - T) \leq w(T - T) + C_1 \mathrm{diam}(T), \quad (*) $$ with $T - T := \left\{u - v : u, v \in T \right\}$, and the hint is to use Gaussian concentration. I have been unable to use this hint, and only end up with a trivial upper bound where $C_1 = \sqrt{n}$, as follows:
$$ h(T - T)^2 = \mathbb{E} \sup_{x \in T - T} \langle g, x \rangle^2 = \mathbb{E} \left( \sup_{x \in T - T} \left\langle g, \frac{x}{\| x \|_2} \right\rangle^2 \| x \|_2^2 \right) \\ \leq \sup_{x \in T - T} \| x \|_2^2 \mathbb{E} \| g \|_2^2 = \mathrm{diam}^2(T) \cdot n, $$ followed by taking square roots.
Question: How does one use Gaussian concentration to show the bound $(*)$? I tried showing that $g \mapsto \sqrt{\sup_{x \in T - T} \langle g, x \rangle^2} - \sup_{y \in T - T} \langle g, y \rangle$ is Lipschitz, but couldn't get anything useful since there is a square root involved.