Answer: $\frac{2\sqrt{3}}{3}$.
The equality holds at $x = \frac{2+\sqrt{3}}{3}$, $y = \frac{2-\sqrt{3}}{3}$, $z = \frac{2}{3}$.
Let us show that $x - y \le \frac{2\sqrt{3}}{3}$.

Solution 1: Problem conditions yield
$$x + y = 2 - z, \quad xy = 1 - z(x + y) = 1 - z(2 - z) = (z - 1)^2$$
Therefore,
$$(x - y)^2 = (x + y)^2 - 4xy = (2 - z)^2 - 4(z - 1)^2 = z(4 - 3z)$$
and consequently $|x - y| = \sqrt{z(4 - 3z)}$. Since by AM-GM inequality $3z(4 - 3z) \le 4$ we can complete the proof: $x - y \le |x - y| \le 2\sqrt{3}/3$.

Solution 2: Problem conditions yield $(x-y)^2 + (y-z)^2 + (z-x)^2 =$
$2(x^2+y^2+z^2)-2(xy+yz+zx) = 2(x+y+z)^2 - 6(xy+yz+zx) = 2.$
Let us put $a = x-y$ and $b = y-z$. Then $z-x = -a-b$ and hence
$a^2 + b^2 + (-a-b)^2 = 2$ and $a^2 + ab + b^2 = 1$. The last equation
can be interpreted as a second order equation with respect to $b$: $b^2 + ab + (a^2-1) = 0$. Since $a$ and $b$ are real numbers its discriminant $D = a^2 - 4(a^2-1) = 4 - 3a^2 \ge 0$. Hence $a \le |a| \le \frac{2}{\sqrt{3}}$.
Done.