EDITED to clarify and incorporate the comments and improve notation (sorry the original was poor). Thank you Derek Holt and Mesel.
Let $G$ be a finite group acting on a set $\Omega$. A block is a subset $\psi$ of letters, {$\omega_k$}, from $\Omega$ with the property that every $g\in G$ either maps the set $\psi$ into (hence onto) itself or onto an image disjoint from $\psi$. There is no other information about the blocks, including whether there are more than the trivial blocks (singleton sets, empty set or all of $\Omega$). Supposing there are blocks that are non-trivial, let $\Psi$ be the collection $\{\psi\}$ of non-trivial blocks. If there are only trivial blocks ($\Psi$ is empty) there is no difficulty, but then the implication asked for in the problem is vacuous. This is the reason for introducing $\Psi$, if only for context.
Problem 6.4 of Wielandt (Finite Permutation Groups) is the following:
For every pair $\alpha , \beta$ in $\Omega$ form the block $\psi_{\alpha\beta}$ which is the intersection of all blocks $\psi\in\Psi$ that contain both $\alpha$ and $ \beta$. The resulting set of the non-empty intersections are sub-blocks which will be now be named $\Psi$. (We need not refer to the original blocks, except for the assumed existence just mentioned. I do not know if this assumption is a real point or if it is obvious.) These are still blocks, as defined above. It is assumed that these new blocks $\psi_{\alpha\beta}$ contain no non-trivial sub-blocks. Show that the stabilizer of $\alpha$ and $ \beta$, $G_{\alpha\beta}$, is the identity for all pairs $\alpha$ and $ \beta$.
Comment: I do not see any reason that $\psi_{\alpha\beta}$ must be non-empty. This is a possible difficulty. In Professor Holt's answer, it appears necessary to assume the existence of blocks containing certain letters, namely, in the notation given above (and in his very good answer), the blocks $\psi_{\gamma\delta}$ and $\psi_{\alpha\gamma}$. Does the existence of $g\in G_{\alpha\beta}$ such that $\gamma^g=\delta$ imply that blocks $\psi_{\alpha\gamma}$ and $\psi_{\gamma\delta}$ exist? For example, such $g$ might be a simple transposition of $\gamma$ and $\delta$. Example: Consider the putative blocks $\{\alpha,\beta\}$ and $\{\gamma,\delta\}$. Or, for $g=(\gamma,\delta)(\epsilon,\mu)$, the blocks $\{\alpha,\beta\}$, $\{\gamma,\epsilon\}$ and $\{\delta,\mu\}$.
(I apologize if the existence question is obvious and I wasted valuable time.)
Attempt: (This does not lead to a conclusion, hence the request for assistance).
Let $\alpha$,$\beta\in\psi_{\alpha\beta}$ be an ordered pair and consider the orbit $(\alpha,\beta)^G$. This orbit includes only pairs that are found in blocks and contains $o(G)/o(G_{\alpha\beta})$ ordered pairs (orbit-stabilizer theorem). Consider also the orbit of the unique block $\psi_{\alpha\beta}$ containing $\alpha$ and $ \beta$. This contains $o(G)/o(G_{\gamma_{\alpha\beta}})$ blocks. Let the sequence of ordered pairs in the orbit $(\alpha,\beta)^G$ be partitioned according to the blocks the pairs belong to. Note that $G_{\alpha\beta}\subset$$G_{\psi_{\alpha\beta}}$.
Then each block in the orbit of $\psi_{\alpha\beta}$ will contain the same number $o(G_{\psi_{\alpha\beta}})/o(G_{\alpha\beta})$ of ordered pairs. In particular, $o(G_{\alpha\beta})$ divides $o(G_{\psi_{\alpha\beta}})$.
Each subset of ordered pairs in the partition, contained in the same block $\psi_{\alpha\beta}$, comprises (via the components of the ordered pairs) all of the elements $\psi_{\alpha\beta}$ since otherwise the subset, broken into elements, would form a non-trivial sub-block. Thus the number of ordered pairs $o(G_{\psi_{\alpha\beta}})/o(G_{\alpha\beta})$ must be of the form $n(n-1)$ where $n$ is the number of elements in the block.
This seems a surprising result, if correct, because it forces the order of $G$ and certain subgroups to be even. Nevertheless, it may be forced by the given data. Still, I fail to get to the desired conclusion.
Obviously, Professor Holt's solution is beautiful, and I am sure it is the one expected. Did I overthink the problem? Or miss the obvious? Probably both. In either case, many thanks for pulling me out of the weeds.
Here is the exercise as stated in the book - there is no mention of block systems.
For any two different points $\alpha,\beta \in \Omega$, let $\psi_{\alpha \beta}$ be the intersection of all blocks of $G$ which contain $\alpha$ and $\beta$. (By previous result, $\psi_{\alpha \beta}$ is also a block of $G$.) Let all $\psi_{\alpha \beta} \ne \Omega$. Further, let every $\psi_{\alpha \beta}$ contain only trivial proper subblocks. Then $G_{\alpha \beta}=1$.
Here is a solution. Suppose not. Then $\exists g \in G_{\alpha \beta}$ with $\gamma^g = \delta$ for some distinct $\gamma,\delta \in \Omega$. Then $\delta \in \psi_{\alpha \gamma}$, and so by the minimality condition $\psi_{\alpha \gamma} = \psi_{\gamma \delta}$, and similarly $\psi_{ \beta \gamma} = \psi_{\gamma \delta}$, so $\psi_{ \alpha \beta} = \psi_{\gamma \delta}$. Now, using similar arguments on the other points in $\Omega$, we get $\psi_{\alpha \beta} = \Omega$, contradiction.