Consider the random variable $Z_{\mathcal S} := \sup_{A \in \mathcal S}|X_A|$, where $Z_A:=\sum_{i \in A}X_i$. Note that $Z_A \sim N(0,k\sigma^2)$ and $Z < \infty$ always. Moreover, it is a standard computation that
$\mathbb E Z_{\mathcal S} = \mathbb E \sup_{A \in \mathcal S}|X_A| \le \sqrt{2k\log|\mathcal S|} < \infty$, and $\sigma_{\mathcal S}^2 := \sup_{A \in \mathcal S}\mathbb E|X_A|^2 = k < \infty$. Therefore, noting that $Z_{\mathcal S} \ge \sup_{A \in \mathcal S}X_A$, the [Borell-TIS ienquality][1] gives

$$
\begin{split}
\forall u \ge 0,\; \mathbb P\left(\sup_{A \in \mathcal S}X_A \ge \sqrt{2k\log|\mathcal S|} + u\right) &\le \mathbb P(Z_{\mathcal S} \ge \sqrt{2k\log|\mathcal S|} + u)\\
&\le  \mathbb P(Z_{\mathcal S} \ge \mathbb EZ_{\mathcal S} + u)\\
&\le \exp(-u^2/(2\sigma_{\mathcal S}^2)) = \exp(-u^2/(2k)).
\end{split}
$$

To make things more interpretable, we do the the change of variable $t:=u/\sqrt{2k}$ to get
$$
\mathbb P\left(\sup_{A \in \mathcal S}X_A \le \sqrt{2k}(\sqrt{\log|\mathcal S|} + t)\right) \le e^{-t^2},\forall t \ge 0.
$$


  [1]: https://en.wikipedia.org/wiki/Borell%E2%80%93TIS_inequality