The answer is $\tfrac{1}{3}$. To see that $C = \tfrac{1}{3}$ is valid simply observe \[\sum_{k = 1}^{n} x_k^2(x_k - x_{k - 1}) = \sum_{k = 1}^{n} \int_{x_{k - 1}}^{x_k} x_k^2 \; dx > \sum_{k = 1}^{n} \int_{x_{k - 1}}^{x_k} x^2 \; dx = \int_0^1 x^2 \; dx = \frac{1}{3}.\]To see that $\tfrac{1}{3}$ is optimal, consider the sequence $(x_0, x_1, \dots, x_n) = (\tfrac{0}{n}, \tfrac{1}{n}, \dots, \tfrac{n}{n})$ as $n \to \infty$. Alternatively, to show the inequality for $C = \tfrac{1}{3}$ we may use the estimate $b^2(b - a) > \tfrac{1}{3}(b^3 - a^3)$ for $b > a > 0$.

I gave two solutions.

That was very nice and logical sym2012

Write \[\sum_k x_k^2(x_k-x_{k-1}) = \sum_k x_k^2\int_{x_{k-1}}^{x_k}1\,du = \int_0^1 \left(\min \{x_k\mid x_k\ge u\}\right)^2\,du > \int_0^1 u^2\,du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3},\]where inequality holds because we cannot have $\min \{x_k\mid x_k\ge u\}=u$ for all $u$, since there are only finitely many $x_k$. To demonstrate that $C=\tfrac{1}{3}$ is maximal, take $x_k = \frac{k}{n}$ for all $k$ and pick increasingly larger $n$, which works because \[\sum_k \frac{k^2}{n^2}\cdot\frac{1}{n} = \frac{\sum_k k^2}{n^3} = \frac{n(n+1)(2n+1)}{6n^3}\]approaches $\frac{1}{3}$ as $n\to\infty$.