Are all connected manifolds homogeneous A topological space $X$ is called homogeneous, if for every two points $x,y \in X$ there exists a homeomorphism $\phi : X \rightarrow X$ s.t. $\phi(x) = y$.
It is not hard to prove that all connected 2-manifolds are homogeneous. The proof basically comes down to the fact that if $D$ is the open disk in $\mathbb{R}^2$ then for every $x,y \in D$ there exists a homeomorphism $\phi : \bar{D} \rightarrow \bar{D}$ such that $\phi(x) = y$, and $\phi \vert_{\partial D}$ is the identity.
Is it true that a general connected manifold is homogeneous?
 A: Yes, any connected topological manifold $X$ of arbitrary dimension $n$ is homogeneous .     
1) The crucial lemma is that given two points $a,b\in \mathbb B^{\circ}$  in the interior of a  closed ball  $\mathbb B \subset \mathbb R^n$, there exists a homeomorphism $f: \mathbb B\to \mathbb B$ which is the identity on $\partial \mathbb B$ and such that $f(a)=b$.    
2) It then follows that if you fix any point $x_0 \in X$, then  the set of points $y\in X$ that can be written $y=F(x_0)$ for some homeomorphism $F:X\to X$ is both open and closed, hence is equal to $X$.
Hence $X$ is homogeneous.
(By the way, an obvious modification of the proof shows that the analogous result is also true for a differential manifold: its diffeomorphisms  act transitively on  the manifold)
Edit: a fishy image
Let me give a physical model which might help visualize the lemma in 1) ( a totally rigorous and amazingly crisp proof is given in t.b.'s great comment).
Imagine you have a spherical fishbowl completely filled with water and a goldfish sitting somewhere in it.
The lemma says that you can send the goldfish to any preassigned place in the bowl by  skilfully (!) shaking the bowl.
