Let $x = \frac{2}{3}a$. Then we have $|a| \le \frac{1}{3}$. Therefore (B) and (D) are excluded. By symmetry, (C) is also excluded. Then only (A) can be correct.

In general, for any $k \in \mathbb{R}$, let $x = \frac{1}{2}ka$. Then the original inequality becomes
$$|a| \cdot |k-1| + \frac{3}{2} |a| \cdot \left|k - \frac{4}{3}\right| \ge |a|^2.$$
This is equivalent to
$$|a| \le |k-1| + \frac{3}{2} \left|k - \frac{4}{3}\right|.$$
We have
$$|k-1| + \frac{3}{2} \left|k - \frac{4}{3}\right| = \begin{cases} \frac{5}{2}k - 3, & k \ge \frac{4}{3}, \\ 1 - \frac{1}{2}k, & 1 \le k < \frac{4}{3}, \\ 3 - \frac{5}{2}k, & k < 1. \end{cases}$$
So
$$\min_{k \in \mathbb{R}} \left\{ |k-1| + \frac{3}{2} \left| k - \frac{4}{3} \right| \right\} = \frac{1}{3}.$$
The inequality is reduced to $|a| \le \frac{1}{3}$. Answer: A.