If $G$ acts transitively and $\Gamma \subseteq \Omega$ is not a block, then each pair of points could be separated

Let $G$ act transitively on $\Omega$. A subset $\Delta \subseteq \Omega$ is called a block if for each $x \in G$ either $\Delta^x \cap \Delta = \emptyset$ or $\Delta^x = \Delta$. If $\Gamma \subseteq \Omega$ is not a block, and $\alpha, \beta \in \Omega$ are arbitrary, does there exists some $g \in G$ such that either
$$
\alpha \in \Gamma^g \mbox{ and } \beta \notin \Gamma^g
$$
or
$$
\alpha \notin \Gamma^g \mbox{ and } \beta \in \Gamma^g.
$$

I know that if $\Gamma$ is not a block I can find such an pair of points, but I want to show that it holds for each pair. Guess somehow it has to do with transitivity?

Solutions Collecting From Web of "If $G$ acts transitively and $\Gamma \subseteq \Omega$ is not a block, then each pair of points could be separated"

No, that is not true in general. For a counterexample, let $|\Omega|=6$ and $G = S_2 \wr S_3$. Then $G$ is imprimitive with block system consisting of three blocks $\Delta_1= \{\alpha,\beta\}$, $\Delta_2,\Delta_3$ of size $2$.

Let $\Delta = \Delta_1 \cup \Delta_2$. Then $\Delta$ is not a block, but the separation condition does not hold for $\alpha,\beta$.

I told you in a comment to your previous post that the exercise in Dixon & Mortimer that asks you to prove this is wrong.