Source: Vershynin, High Dimensional Probability, Exercise 7.6.1.
Let $T\subset R^n$ and $g\sim N(0,I_n)$. We define the 1-norm and 2-norm Gaussian widths as follows:
$h(T)^2:=\mathbb E\sup_{t\in T} \langle g,t \rangle^2$ and $w(T):=\mathbb E\sup_{t\in T} \langle g,t \rangle$. I want to prove the following inequality:
$$h(T-T)\le w(T-T)+C\text{diam}(T), $$
where $T-T:=\{x-y:x,y\in T\}.$
The hint says one can use the Gaussian concentration for this bound. I think he meant 
I have thought about taking $f(y):=\sup_{x\in T-T}\langle y,x \rangle$, but I feel like we can't get it straightforwardly. Did I miss anything?
