Consider the relation defined by $0 < |x - y| \le 1$.
We need to show that this relation is symmetric but not transitive. A relation is symmetric if whenever $(a, b)$ is in the relation, $(b, a)$ is also in the relation. Since $|x-y| = |y-x|$, if $0 < |x-y| \le 1$, then $0 < |y-x| \le 1$. Therefore, if $(a, b)$ is in the relation, so is $(b, a)$, so the relation is symmetric.

A relation is transitive if whenever $(a, b)$ and $(b, c)$ are in the relation, $(a, c)$ is also in the relation. Let $x = 0.2, y = 0.9, z = 1.5$. Then $|x - y| = |0.2 - 0.9| = |-0.7| = 0.7$, and $|y - z| = |0.9 - 1.5| = |-0.6| = 0.6$. So $0 < |x - y| \le 1$ and $0 < |y - z| \le 1$. However, $|x - z| = |0.2 - 1.5| = |-1.3| = 1.3$, which is not less than or equal to $1$. So $(x, z)$ is not in the relation. Therefore, the relation is not transitive. So, option B is not correct.