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Starting with the definition of variance for a continuous variable X,
\begin{align*} \sigma^2 & = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) dx\\ & \ge \int_{-\infty}^{\mu-a} (x - \mu)^2 f(x) dx + \int_{\mu + a}^{\infty} (x - \mu)^2 f(x) dx\\ & \ge \int_{-\infty}^{\mu-a} a^2 f(x) dx + \int_{\mu + a}^{\infty} a^2 f(x) dx\\ & = a^2 \left ( \int_{-\infty}^{\mu-a} f(x) dx + \int_{\mu + a}^{\infty} f(x) dx \right )\\ & = a^2 P( X \le \mu - a \text{ or } X \ge \mu + a )\\ & = a^2 P( \big | X - \mu \big | \ge a) \end{align*}
Dividing by \(a^2\) and taking the complement gives the result.
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