It is really easy to explain if I were able to draw it fofr you. Indeed, you should draw a picture yourself, or get a bicycle inner tube, or even a donut, for what follows.
Take a donut draw a circle round the 'crown' (i.e. on the top of the donut, where the chocolate frosting is) going round the hole. There's one S^1, now pick a point on the circle as a base point. How do I specify the other points on the donut? I go round this circle some way and then travel on a second circle that goes off at right angles and loops round the body of the donut. There, the Torus T^1 is the same as S^1xS^1
The torus is also the same as the unit square with opposite sides glued together, which also shows that it is like S^1xS^1, since S^1 is the same as the unit interval with the end points identified.
Take S^2, remove an open disc, then you can imagine spreading it out by putting your fingers into the hole and pulling, and you'll get a disc, necessarily closed if we are only thinking of homeomorphism (homeomorphisms must be 'undoable' in this sense, homotopies need not be, thus a sphere with any point, open disc, or closed disc, removed is homotopic to a point.
An open disk is something homeomorphic to the set of points {(x,y) : x^2+y^2<1}, or better, it is well, a disc without its boundary (thus making it open).
Removing an open disc is punching a hole in the sphere, but leaving the boundary of the hole there (so what remains is closed).
These thigns really are a lot easier with a pad of paper to draw them on.
here, try this
http://www.answers.com/topic/torus
(googles third hit on searching for 'torus homeomorphic product circles', by the way), about a 1/3 of the way down it draws a picture of the torus and the two embedded circles idea.