If $X_i, ...., X_d$ are $b$-subGaussians, for any $t \in \mathbb{R}^{+}$:
$$ \mathbb{E}e^{tX} \leq \mathbb{E} e^{t^2 b^2 / 2} $$
Then for $Z = \max_{i=1}^n X_i$ I think I can write (since it is one of those subGaussian variables):
$$
\mathbb{E}e^{tZ} \leq \mathbb{E} e^{t^2 b^2 / 2}
$$
which proves that it is subGaussian with the same parameter.
Am I correct?
Update: Actually now I think I am wrong.
$$ \mathbb{E}e^{t\max_i X_i} = \mathbb{E}\max_i e^{tX_i} $$
Now in order continue the argument I need to be able to bring the max outside the expectation. However Jensen's inequality tell me that: $$ \max (\mathbb{E}Y) \leq \mathbb{E} \max Y $$
Which is the reverse of the direction I need.
No, it's not true. As your reference states, a necessary condition for $Z$ to be subgaussian is that $\mathbb{E}Z = 0$. However, that is not true in general.
For a simple example that's easy to compute explicitly, let $X_i$ be independent Rademacher distributed. Then $Z=-1$ with probability $2^{-n}$ and $Z=1$ with probability $1-2^{-n}$, so $\mathbb{E}Z = 1-2^{1-n} >0$.