A picture proof that sets in the ordered Mostowski model have a least support In Jech's book on the Axiom of Choice he presents in chapter 4.5 the ordered Mostowski model $\mathcal{V}$. One useful fact about this model is that every $x \in \mathcal{V}$ has a least support. The following lemma is used to show this:

Lemma: If $E_1$ is a support of $x$ and $E_2$ is a support, then $E_1 \cap E_2$ is a support of $x$.

About the proof of this lemma Jech only comments: "the proof is based on the following fact, which can be proved
more easily by drawing a picture than in writing, and thus we leave it to
reader's imagination: Let $E_1$ and $E_2$ be finite subsets of $A$, $E = E_1 \cap E_2$. If $\pi \in G$ is such that $\pi(a) = a$ for each $a \in E$, then we can find a finite number of permutations $\pi_1\rho_1, \dots, \pi_n\rho_n$ such that $\pi_i \in \text{fix}(E_1)$ and $\rho_i \in \text{fix}(E_2)$ for each $i = 1, \dots, n$, and $\pi = \pi_1\rho_1 \dots \pi_n\rho_n$.
What is the proof and picture that Jech is thinking of?
 A: The point is that this is something that once you visualise, is easy to understand how a proof should look like, but nevertheless, writing one in full detail is an excruciatingly long and borderline unnecessary torture (which, by the way, is illegal under the Geneva Conventions, so by asking this very question you're risking breaking international law).
Here is the gist. Suppose first that $E_1\cap E_2=\varnothing$, this means that we should be able to write any permutation in our group as a product of permutations from $\DeclareMathOperator{\fix}{fix}\fix(E_1)\cup\fix(E_2)$.
Of course, any permutation that does not require moving anything from $E_1\cup E_2$ is already doing the job, in fact even moving only points from at most one of them is already its own decomposition. So we might as well assume that we move something from each of $E_1$ and $E_2$.
Say that we moved $a\in E_1$ and $b\in E_2$. So why not just assume for a start that $E_1=\{a\}$ and $E_2=\{b\}$? That's simpler. Let me call our permutation $\pi$. And without loss of generality $a<b$.
The more difficult case is that when $a<b<\pi(a)<\pi(b)$, and equivalently when $\pi(b)<a$. And by understanding this one, the simpler case where $a<\pi(a)<b<\pi(b)$ is also understood.
Let $U$ be a large enough interval containing $a,b,\pi(a),\pi(b)$. And now partition $U$ into open intervals: $U_a$ contains $a$, $U_b$ contains the other three points, and $U_c$ is a third intervals which lies between $U_a$ and $U_b$. Note that we use the fact that the rationals are totally disconnected here.
First we move with some $\pi_0$ the interval $U_b$ to $U_c$ and fix $U_a$ pointwise. This is an order isomorphism and it lies in $\fix(\{a\})$. Next move $a$ to $\pi_0(\pi(a))$, and $\pi_0(b)$ to $\pi_0(\pi(b))$. But we can do better, there is some $\pi_1\in\fix(\{b\})$ which will satisfy that $\pi_0^{-1}\circ\pi_1$ acts like $\pi$ on a large enough interval around $a$ and $b$. 
Finally, we can correct the necessary values of $\pi_0^{-1}\pi_1\pi_0$ by some permutation moving only things outside that interval.

The more complicated your $E_1$ and $E_2$ are, and the more $\pi$ moves them between the different intervals given by the points in each set, the more complicated the above idea becomes. But ultimately, this is something that follows from the fact that partial automorphisms can be extended to automorphisms. So we move them "slowly" and "carefully".
