Connected manifold implies homogeneous

Statement
Any connected manifold is homogeneous, viz given a connected manifold and two points in it, there is a self-homeomorphism of the manifold that takes the first point to the second.

Proof outline
The proof involves three steps:


 * We show that given any two points inside a closed disk in Euclidean space, there is a homeomorphism of the disc that takes one point to the other and is identity on the boundary.
 * We use the gluing lemma for closed subsets to show that if two points lie inside a Euclidean open subset of the manifold, then there is a homeomorphism of the manifold that takes one point to the other. To apply the gluing lemma for closed subsets, we use the fact that the disc is compact, and that the manifold is Hausdorff, and hence the disc is a closed subset of the manifold.
 * We finally use the fact that the manifold is locally Euclidean to show that the orbit of any point under the action of the self-homeomorphisms is both open and closed, and then use connectedness of the manifold to show that it is the whole manifold.

A further abstraction
The proof outline can be abstracted by defining the notion of a compactly homogeneous space -- a space in which given any two points, there is a homeomorphism between them that fixes the complement of a compact set. The above proof then generalizes to the fact that a connected homogeneous space in which every point is contained in a compactly homogeneous open set, is itself compact homogeneous.

A Proof by Contraposition
Suppose there exists a closed, connected n-manifold N which is not homogeneous. Then there exist points a and b in N such that for every phi in Aut(N), phi(a) is not equal to b. Consider the orbit of a induced by Aut(N), denoted Aut(N)(a); since N is locally homeomorphic to R^n, Aut(N)(a) is open, and for the same reason, it is also closed: Let B be an open ball with center phi(a) contained in a neighborhood of phi(a) homeomorphic to R^n; we know that D^n is trivially homogeneous (by constructing a homeomorphism  chi to R^n that sends the boundary to infinity, then composing chi with a translation between two desired points, and finally composing this with the inverse of chi, which yields an automorphism on D^n between any two points), so for every x in B, there exists psi in Aut(N) such that psi(phi(a))=x, but psi composed with phi is again in Aut(N), so x is in Aut(N)(a), therefore for every b in Aut(N)(a), there exists a neighborhood of b contained in Aut(N)(a), so Aut(N)(a) is open. To show that Aut(N)(a) is closed, consider the complement of Aut(N)(a) in N: let c be in the complement, then there exists an open ball B' with center c contained in a neighborhood of c homeomorphic to R^n, and for every x in B', there exists rho in Aut(N) such that rho(c)=x, so x cannot be in Aut(N)(a), since if there exists zeta in Aut(N) such that zeta(a)=x, then (rho^(-1))(zeta(a))=c, but zeta composed with rho inverse is in Aut(N), so c is in Aut(N)(a), a contradiction, therefore the complement of Aut(N)(a) in N is open, so Aut(N)(a) is closed; therefore Aut(N)(a) is clopen. Note that Aut(N)(a) is a nontrivial, proper subset of N, since by the trivial automorphism a is in Aut(N)(a), and b is by definition not in Aut(N)(a), but then there exists a nontrivial, proper, clopen subset of N, which implies that N isn't connected, a contradiction, therefore N is homogeneous.