Why does a sphere become homeomorphic to a plane if we remove a point from it? What is the intuition behind the fact that removing a point from a sphere makes it homeomorphic to a plane?
 A: For a physical intuition:
Imagine a sphere made of extremely elastic rubber. 
If you have a small hole in it, you can squeeze your fingers in than expand that hole ever more, not tearing the fabric. 
At some point the boundary of the initial hole will form about a disk larger than the initial sphere. Keep tearing and tearing you'll have an ever larger disk and 'eventually' a plane.  
If you do not have a hole, you do not have a place to start the process. 
For something still intuitive, bot more mathematical:
Imagine the sphere lying on the origin of the plane, and the missing point is the North Pole. From the North Pole of the sphere, project each point (other than the North Pole) to the plane. That is form the line through the North Pole and that point and find the intersection with the plane. 
It is not hard to see this map is a hoemomorphism. 
A: That a punctured sphere is homeomorphic to the plane can be proven by producing an explicit homeomorphism, like the stereographic projection described in quid's answer. That's the easy part.
It is much more difficult to prove that a "perfect" sphere $S^2$ is not homeomorphic to the plane ${\mathbb R}^2$. We cannot look at all maps $f:\>S^2\to{\mathbb R}^2$ individually and detect some flaw in each of them.
The mathematical field of topology has set up a catalogue of properties, or invariants, of spaces that remain unchanged under homeomorphisms. If $S^2$ has one such property, and ${\mathbb R}^2$ does not, then we are sure that the two spaces are not homeomorphic. This is the case for the property "compactness", which unfortunately seems out of bounds for you.
Therefore let me give the following intuitive argument for the claim that $S^2$ and ${\mathbb R}^2$ are not homeomorphic: If you hammer a nail into $S^2$ you can still contract any loop on $S^2$ to a point. But if you hammer a nail into ${\mathbb R}^2$ some loops cannot be contracted to a point any more, namely the ones going around the nail.
A: Probably part of your confusion comes from not fully appreciating what homeomorphisms can do. One should think of topological spaces as being extremely malleable and flexible under continuous maps. Homeomorphisms can exploit that to a great degree and seem counterintuitive sometimes. If you are new to topology then you are probably bringing a lot of intuition from geometry and takes effort to release yourself from that. Of course geometrically the punctured sphere and the plane are far apart and so it feels strange at first that they are homeomorphic. 
I like the answer by quid. One of my intuitive crutches here is to, instead of puncturing it with one point, puncture a massive hole in it: remove a hemisphere. Now what you're left with is much more obviously disk-like. It's not as hard to believe that a disk and the plane are homeomorphic. But again, part of this boils down to just "believing" that you can do all these stretches and bends while still remaining homeomorphic. It's actually helpful to try to construct homeomoprhisms explicitly on a few simple examples. You quickly start to see that you can get away with a lot!
