Separation of finite sets in homogeneous spaces by homeomorphisms Call a topological space $X$ flexible, if for each finite set $A \subset X$ there exists a homeomorphism $f: X \rightarrow X$ such that $A \cap f(A) = \emptyset$.
(Certainly not a good name, and by far not standard, but for the purpose of this question it might suffice.)
Let $X$ be an infinite, T2, homogeneous topological space (i.e., for all $x, y \in X$ there is a homeomorphism $f: X \rightarrow X$ such that $f(x) = y$).
Is $X$ flexible?
Notes

*

*Of course, a flexible, non-empty space is infinite and must provide a certain amount of homeomorphisms.
For instance, if it is rigid (i.e, the identity is the only homeomorphism), it can't be flexible. Therefore, it makes sense to restrict to homogeneous spaces.

*Considering "typical" homogeneous spaces as $\mathbb{R}^n$, the answer seems so obviously to be "yes". However, I couldn't prove the above in general, not even for two-element sets $A$.

*It is not difficult to prove that $X$ is flexible, if $X$ is infinite and at least one of the following conditions holds: 
a) $X$ is the underlying space of a topological group 
b) $X$ is a product with (at least) one factor flexible (this might indicate how weak flexible is)
c) $X$ is n-homogeneous for all $n \in \mathbb{N}$ 
d) $X$ is strongly locally homogeneous, T2 and contains no isolated points (in particular, if $X$ is a manifold) 
e) $X$ is uniquely homogeneous 
(The notations in c), d) and e) are the standard ones, see for instance here.)

*My assumption is that the answer is "yes". Perhaps, the proof is more combinatorial (eg. Ramsey theory) rather than topological?
Or even with some trivial argument, which I just didn't notice?

*The pseudo-arc is a standard example of a homogeneous, not strongly locally homogeneous, space. I'm not very familiar with it. Embarrassingly, I don't know, whether it is flexible or not. Perhaps, it provides a counter-example?

*[edit: I just deleted 6. (and my two related comments below), since after some further consideration it no longer make sense.]

*Perhaps the T2 requirement in the prerequisite is superfluous? I also don't know of a non-T2 counterexample.

 A: This is in fact automatic for purely combinatorial reasons.  By a theorem of Neumann, a group cannot be covered by finitely many cosets of subgroups of infinite index (see https://mathoverflow.net/questions/17396/can-a-group-be-a-finite-union-of-left-cosets-of-infinite-index-subgroups).  As a corollary, if a group $G$ acts transitively on an infinite set $X$, then for any finite $A,B\subset X$ there exists $g\in G$ such that $gA\cap B=\emptyset$.  Indeed, if no such $g$ existed, that would mean exactly that $G$ is covered by the finitely cosets of the stabilizer subgroups of each element of $A$ which map them to each element of $B$.  These stabilizer subgroups all have infinite index because $G$ acts transitively and $X$ is infinite, so this is impossible.
Here is a direct proof of that corollary (this is just what you get by translating Neumann's proof into the language of group actions).  We use induction on $|A|$, the base case $|A|=0$ being trivial.  Now suppose $|A|>0$ and fix $a\in A$.  Pick $h\in G$ such that $h(a)\not\in B$, and also for each $b\in B$ pick $g_b\in G$ such that $g_b(a)=b$.  Now apply the induction hypothesis to the sets $A'=A\setminus\{a\}$ and $B'=B\cup\bigcup_{b\in B}g_bh^{-1}B$ to obtain $g\in G$ such that $gA'\cap B'=\emptyset$.  If $g(a)\not\in B$ then we have $gA\cap B=\emptyset$ and are done.  If $g(a)\in B$, let $b=g(a)$ and observe that $hg_b^{-1}g(a)=h(a)\not\in B$ and also for each $a'\in A'$ we have $hg_b^{-1}g(a')\not\in B$ since $g(a')\not\in g_bh^{-1}B$.  Thus $hg_b^{-1}gA\cap B=\emptyset$ and $hg_b^{-1}g$ is our desired element of $G$.
A: This answer currently has a fatal flaw.
Your supposition 4 is probably correct, in that the proof runs through infinitary combinatorics.  I use topology below, but purely instrumentally; with a little more cleverness, it probably could be eliminated.  This also implies your supposition 7.  Now, on to the proof.
Let $X$ be an set on which $G$ acts transitively and (w/oLoG) faithfully.  (For example, let $X$ be an infinite topological space and $G=\mathrm{Homeo}(X)$.)  For any finite $A,B\subseteq X$, I claim that either there exists $g\in G$ such that $gA\cap B=\emptyset$, or $X$ is finite. To reduce to your question, take $A=B$ in an infinite $X$.
Fix $x\in X$ and pick any $\{g_a\}_{a\in A\cup B}$ such that $g_ax=a$ (for all $a\in A\cup B$).  Note that $$\{g:gA\cap B=\emptyset\}=\bigcap_{a\in A,b\in B}{\{g:ga\neq b\}}=\bigcap_{a\in A,b\in B}{g_b(G\setminus\mathrm{Stab}(x))g_a^{-1}}\tag{1}$$
Now, give $X$ the coarsest topology such that $X$ is $T_1$: the cofinite topology.  (If $X$ is finite, this is discrete.)  Place on $G$ the initial topology induced by the action on each point of $X$; that is, $$\{\{g:gy\in U\}:y\in X,U\text{ open in }X\}$$ is a basis (flaw: this should be a subbasis) for this topology.  $G$ is not necessarily a topological group under this topology, but the following nice properties still hold:

*

*For any $l,r\in G$, the maps $p\mapsto lp$ and $p\mapsto pr$ are homeomorphisms.  This follows from the explicit $G$-invariance of our basis.

*The set $\mathrm{Stab}(x)$ is closed.  This follows immediately from the definition.

*If the set $\mathrm{Stab}(x)$ is not nowhere dense in $G$ (i.e. has nontrivial interior), then $X$ is finite.

To see the last claim, let $U$ be open and $y\in X$ such that $$\{g:gy\in U\}\subseteq\mathrm{Stab}(X)$$  Let $K=X\setminus U$, pick any $u\in U$, and, for each $k\in K$, choose $g_k\in G$ such that $g_ku=k$.  I claim that $$G=\mathrm{Stab}(x)\cup\bigcup_{k\in K}{g_k\mathrm{Stab}(x)}$$  For, if $g\in G$, then either

*

*$gy\in U$, so that $g\in\mathrm{Stab}(x)$, or

*$gy\in K$, so that $g_{gy}^{-1}gy=u\in U$ and $g_{gy}^{-1}g\in\mathrm{Stab}(x)$.

Thus $$|X|=|G/\mathrm{Stab}(x)|\leq|K|+1<\aleph_0$$ since $U$ is cofinite.
Now, $G$ with this topology need not be Baire.  Nevertheless, in any topological space, a finite intersection of dense open sets is dense.  Thus (1) is dense, and, in particular, nonempty.
A: To complete the picture I would like to add the following lemma, which is a trivial consequence of Eric Wofsey's proof:
Let the group $G$ be acting on the set $X$. Then it holds: 
For  all $A, B$ finite subsets of $X$ there is a $g \in G$ such that $gA \cap B = \emptyset$,
if and only if each orbit is infinite.
( "$\Rightarrow$" is obvious. "$\Leftarrow$": Follows by the same proof as above: $h$ can be picked, since the orbit of $a$ is infinite.
Instead of $b \in B$ one has to consider $b \in B \cap \mathrm{orbit}(a)$.)
As a corollary we have that a topological space is flexible (as defined at the top), if and only if each homogeneity component is infinite.
