To determine the greatest real number $C$ such that for every positive integer $n \geq 2$, there exist $x_1, x_2, \ldots, x_n \in [-1, 1]$ satisfying
$$\prod_{1 \le i < j \le n} (x_i - x_j) \geq C^{\frac{n(n-1)}{2}},$$
we consider the example where $x_i = \cos\left(\frac{i\pi}{n}\right)$ for $i = 1, 2, \ldots, n$.

For this choice, the product $\prod_{1 \le i < j \le n} (x_i - x_j)$ can be analyzed using properties of Chebyshev polynomials. Specifically, the roots of the Chebyshev polynomial of degree $n$ are given by $\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k = 1, 2, \ldots, n$. The difference between any two such roots can be expressed in terms of sine functions:
$$x_j - x_i = 2 \sin\left(\frac{(i-j)\pi}{2n}\right) \sin\left(\frac{(i+j)\pi}{2n}\right).$$

The logarithm of the product of these differences can be approximated by considering the average value of $\ln \left| \sin(x) \right|$ over the interval $[0, 2\pi]$, which is $\ln \left(\frac{1}{2}\right)$.

Thus, the product $\prod_{1 \le i < j \le n} (x_i - x_j)$ is approximately $\left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}}$.

Therefore, the greatest real number $C$ satisfying the given inequality for all $n \geq 2$ is $\frac{1}{2}$.

The answer is: \boxed{\frac{1}{2}}.