= Solution
Use the convention $\mathbb E e^{sX}\leq e^{\sigma^2s^2/2}$ for the centered <sub-Gaussian random variable>. Applying the two-sided <Chernoff bound> to $X$ gives $\mathbb P(|X|>x)\leq2e^{-x^2/(2\sigma^2)}$. Since $\mathbb E X^2\geq0$, this already supplies the requested bound:
$$
\boxed{\mathbb P(X^2-\mathbb E X^2>t)\leq\min\{1,2e^{-t/(2\sigma^2)}\},\qquad t>0.}
$$
For averages of <independent random variables>, the stronger useful statement is that the <centered square of a sub-Gaussian random variable> is a <sub-exponential random variable>. The <Bernstein bound for independent sub-exponential variables> then give $\mathbb P(n^{-1}\sum_i(X_i^2-\mathbb E X_i^2)>t)\leq2\exp[-c n\min\{t^2/\sigma^4,t/\sigma^2\}]$ for a universal positive $c$. These are the concentration theorems used for quadratic <scan statistics>.
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