The similar and more general question is asked here, whose setting is random vectors.
Let $X$ be $\sigma$-sub-Gaussian and $f$ is a Lipschitz function w.r.t. constant $L$. How to prove can the subgaussian moment of $f(X)$ be bounded in terms of $\sigma$ and $L$?
It might not work on random vectors but I think it works on random variable.
$\newcommand{\si}{\sigma}\newcommand\R{\mathbb R}$One of mutually equivalent definitions of a ($\si$)-sub-Gaussian random variable (r.v.) $X$ is as follows: \begin{equation*} Ee^{X^2/\si^2}\le2. \tag{1}\label{1} \end{equation*}
It is now clear that the answer to the question is no. Indeed, suppose that $X\sim N(0,1)$, so that $X$ is $\si$-sub-Gaussian for some universal positive real constant $\si$. For any real $a$ and $x$, let now $f_a(x):=x+a$, so that $f_a$ is $L$-Lipschitz for $L=1$. Then for any real $b>0$ we have $Ee^{f_a(X)^2/b^2}\to\infty$ as $a\to\infty$, so that the sub-Gaussian norm of $f_a(X)$ cannot be bounded in terms of $\si$ and $L$.
However, the sub-Gaussian norm of $f(X)-Ef(X)$ can be bounded in terms of $\si$ and $L$, as follows. Let $Y$ be an independent copy of $X$. Then, for any real $c>0$, by Jensen's inequality for the convex function $u\mapsto\exp\frac{(t-u)^2}{c^2}$, \begin{equation*} \begin{aligned} E\exp\frac{(f(X)-Ef(X))^2}{c^2}&=E\exp\frac{(f(X)-Ef(Y))^2}{c^2} \\ &=\int_\R P(X\in dx)\exp\frac{(f(x)-Ef(Y))^2}{c^2} \\ &\le\int_\R P(X\in dx)E\exp\frac{(f(x)-f(Y))^2}{c^2} \\ &=E\exp\frac{(f(X)-f(Y))^2}{c^2} \\ &\le E\exp\frac{L^2(X-Y)^2}{c^2} \\ &\le E\exp\frac{2L^2X^2+2L^2Y^2}{c^2} \\ &=\Big(E\exp\frac{2L^2X^2}{c^2}\Big)^2 \\ &\le E\exp\frac{4L^2X^2}{c^2}\le2 \end{aligned} \end{equation*} by \eqref{1} if \begin{equation*} c=2L\si. \end{equation*} So, the sub-Gaussian norm of $f(X)-Ef(X)$ is $\le2L\si$, if $\si$ is the sub-Gaussian norm of $X$. $\quad\Box$
