We can view Sn as a (closed) subset of Rn so it automatically inherits hausdorfness and second countabiltiy. To show it is a manifold, you need to cover it by charts. For example, lets take S1. The first chart you would take would be the open upper hemisphere, ie to the top half of the circle minus the points (0, 1) and (0-1). We need to do this to ensure this set is open in the subspace topology. Now as your coordinate function, just take the map f (x,y) = x. This just projects those points down onto the real line, showing this open set is homeomorphic to R. Then you just cover the rest of the circle using similar ideas ( hint, you will need 3 more patches in total). Then you show that the transition functions are smooth, which isnt hard to check. To extend this to S^n isnt too