[proofplan]
We verify the sub-Gaussian MGF condition for the centred linear combination $\sum_{i=1}^n \gamma_i(W_i - \mathbb{E}[W_i])$. Independence of $W_1, \ldots, W_n$ allows us to factorise the joint moment generating function into a product of individual MGFs. Each factor is then bounded using the sub-Gaussian condition on $W_i$ with the rescaled parameter $\alpha \gamma_i$. Multiplying the bounds and collecting exponents yields the sub-Gaussian condition with the stated parameter.
[/proofplan]
[step:Centre the linear combination and factorise its MGF using independence]
The centred version of $\gamma^\top W = \sum_{i=1}^n \gamma_i W_i$ is
\begin{align*}
\gamma^\top W - \mathbb{E}[\gamma^\top W] = \sum_{i=1}^n \gamma_i(W_i - \mathbb{E}[W_i]),
\end{align*}
by linearity of expectation. For any $\alpha \in \mathbb{R}$, consider the moment generating function of this centred sum:
\begin{align*}
\mathbb{E}\!\left[\exp\!\left(\alpha \sum_{i=1}^n \gamma_i(W_i - \mathbb{E}[W_i])\right)\right] = \mathbb{E}\!\left[\prod_{i=1}^n \exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right].
\end{align*}
Since $W_1, \ldots, W_n$ are independent, the random variables $\exp(\alpha \gamma_i(W_i - \mathbb{E}[W_i]))$ for $i = 1, \ldots, n$ are independent (as measurable functions of independent random variables). The expectation of a product of independent random variables equals the product of their expectations:
\begin{align*}
\mathbb{E}\!\left[\prod_{i=1}^n \exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right] = \prod_{i=1}^n \mathbb{E}\!\left[\exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right].
\end{align*}
[guided]
The factorisation step is where the independence hypothesis is consumed. Without independence, the joint MGF would involve the full joint distribution of $(W_1, \ldots, W_n)$, and no factor-by-factor bound would be possible.
The centred version of the linear combination is $\sum_{i=1}^n \gamma_i(W_i - \mathbb{E}[W_i])$. This uses linearity of expectation: $\mathbb{E}[\sum_i \gamma_i W_i] = \sum_i \gamma_i \mathbb{E}[W_i]$.
We write the exponential of a sum as a product of exponentials:
\begin{align*}
\exp\!\left(\alpha \sum_{i=1}^n \gamma_i(W_i - \mathbb{E}[W_i])\right) = \prod_{i=1}^n \exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr).
\end{align*}
Each factor $\exp(\alpha \gamma_i(W_i - \mathbb{E}[W_i]))$ depends on $W_i$ alone. Since $W_1, \ldots, W_n$ are independent, these factors are independent random variables. For independent random variables $X_1, \ldots, X_n$ with $\mathbb{E}[|X_i|] < \infty$, we have $\mathbb{E}[\prod_i X_i] = \prod_i \mathbb{E}[X_i]$. Each factor has finite expectation because the sub-Gaussian condition ensures $\mathbb{E}[e^{\alpha \gamma_i (W_i - \mathbb{E}[W_i])}] \leq e^{(\alpha \gamma_i)^2 \sigma_i^2 / 2} < \infty$. Hence
\begin{align*}
\mathbb{E}\!\left[\prod_{i=1}^n \exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right] = \prod_{i=1}^n \mathbb{E}\!\left[\exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right].
\end{align*}
[/guided]
[/step]
[step:Bound each factor using the sub-Gaussian condition on $W_i$]
For each $i \in \{1, \ldots, n\}$, the random variable $W_i$ is sub-Gaussian with parameter $\sigma_i$, which by definition means
\begin{align*}
\mathbb{E}\!\left[\exp\!\bigl(s(W_i - \mathbb{E}[W_i])\bigr)\right] \leq \exp\!\left(\frac{s^2 \sigma_i^2}{2}\right) \quad \text{for all } s \in \mathbb{R}.
\end{align*}
Setting $s = \alpha \gamma_i$ (which is a real number for any fixed $\alpha \in \mathbb{R}$ and $\gamma_i \in \mathbb{R}$):
\begin{align*}
\mathbb{E}\!\left[\exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right] \leq \exp\!\left(\frac{\alpha^2 \gamma_i^2 \sigma_i^2}{2}\right).
\end{align*}
[/step]
[step:Multiply the bounds and identify the sub-Gaussian parameter]
Taking the product of the bounds from the previous step over $i = 1, \ldots, n$:
\begin{align*}
\prod_{i=1}^n \mathbb{E}\!\left[\exp\!\bigl(\alpha \gamma_i(W_i - \mathbb{E}[W_i])\bigr)\right] \leq \prod_{i=1}^n \exp\!\left(\frac{\alpha^2 \gamma_i^2 \sigma_i^2}{2}\right) = \exp\!\left(\frac{\alpha^2}{2} \sum_{i=1}^n \gamma_i^2 \sigma_i^2\right).
\end{align*}
The equality uses $\prod_i e^{a_i} = e^{\sum_i a_i}$. Combining with the factorisation from the first step:
\begin{align*}
\mathbb{E}\!\left[\exp\!\left(\alpha\!\left(\gamma^\top W - \mathbb{E}[\gamma^\top W]\right)\right)\right] \leq \exp\!\left(\frac{\alpha^2}{2} \sum_{i=1}^n \gamma_i^2 \sigma_i^2\right).
\end{align*}
This holds for all $\alpha \in \mathbb{R}$. Writing $\tilde{\sigma}^2 := \sum_{i=1}^n \gamma_i^2 \sigma_i^2$, the bound becomes $\mathbb{E}[e^{\alpha(\gamma^\top W - \mathbb{E}[\gamma^\top W])}] \leq e^{\alpha^2 \tilde{\sigma}^2 / 2}$, which is exactly the sub-Gaussian condition with parameter
\begin{align*}
\tilde{\sigma} = \left(\sum_{i=1}^n \gamma_i^2 \sigma_i^2\right)^{1/2}.
\end{align*}
Therefore $\gamma^\top W$ is sub-Gaussian with parameter $\left(\sum_{i=1}^n \gamma_i^2 \sigma_i^2\right)^{1/2}$.
[/step]
