In the mentioned paper, they have $\sum_{t=1}^v$ rather than $\sum_{t=0}^v$. Anyhow,
the inequality is incorrect in general. Indeed, without loss of generality, $\mu=0$ and $\sigma=1$. Suppose that (say) the $X_i$'s are just standard normal.
Then, letting $n:=2^{u+1}[\ge4]$ and $S_v:=\sum_{t=1}^v X_t$, we see that $S_{n/2},\dots,S_n$ are jointly normal, and so, they have a bounded joint density. So,
the lhs of the inequality is $1-O(\delta^{n/2+1})=1-o(\delta^2)$ as $\delta\downarrow0$, whereas the rhs is $1-\delta^2/(2+o(1))$.

However, the inequality can be easily fixed as follows, likely without any serious damage to the proof of that theorem in that paper (again assuming $\mu=0$ and $\sigma=1$): In view of Doob's maximal inequality for submartingales and the sub-Gaussian property, the lhs of your inequality (with $\sum_{t=1}^v$ rather than $\sum_{t=0}^v$) is no greater than
\begin{align*}
P(\max_{0\le v\le n}|S_v|>\delta\sqrt{n/2})
&\le
e^{-h\delta\sqrt{n/2}}Ee^{h|S_n|} \\
& \le e^{-h\delta\sqrt{n/2}}(Ee^{hS_n}+Ee^{-hS_n}) \\
& \le 2\exp\{-h\delta\sqrt{n/2}+nh^2/2\}
=2\exp\{-\delta^2/4\},
\end{align*}
where $h:=\delta/\sqrt{2n}$.