Describe subspace topology on the 2- sphere $S^2$ and find base of this topology on $S^2$ Describe subspace topology on the 2- sphere $S^2$ w.r.t to inclusion $S^2 \subset R^3$, where $R^3$  is provided with euclidean topology,  and find basis of this topology on $S^2$
I am new to topology and lot of definitions are mixed in my head. That's why don't blame me too hard, if I write stupid things.
1) I didn't understand precisely, what does in mathematical sense the word "describe" mean? But I did the following:
$S^2 \subset R^3$ and suppose $O \subset R^3$, then by definition of subset topology  $U = S^2\cap O$, where $U \subset R^2$. Because subspace topology is defined as intersection of two sets, that are in Euclidean topology, the same rules works for subset topology.
2)Base of subset topology on $S^2$:
Base B for a topological space X with topology T -  is a collection of open sets in T such that every open set in T can be written as a union of elements of B.
let $O$ be open set in subset topology and $x \in O$. By definition of open set there exist neighbourhood around $x$, s.t  and for every $y \in O$ $||x - y||<\epsilon $ $N_\epsilon(x) \subset O$. We know, that neighbourhoods are open.
We know, that every open set is a union of neighborhoods. The first assumption of base is satisfied. The second is satisfied as well, because intersection of finitely many open set is open set.
Can you please correct my work and give some feedback?
 A: Loosely speaking, when some question asks you to "describe" something means to find some intrinsic property. 
1) Since $S^{2}\subset\mathbb{R}^{3}$ and $O\subset\mathbb{R}^{3}$ (actually $O$ is not a random subset, need to be open, since you are describing the subspace topology of $S^{2}$), then we must have $O\cap S^{2}=U\subset\mathbb{R}^{3}$ (not $\mathbb{R}^{2}$). The fact that $S^{2}$ is locally $\mathbb{R}^{2}$ doesn't mean that its subsets belongs to it. So now you have a description of the open sets of $S^{2}$.
2) Some corrections in your definition: You must have that $B$ is countable, by definition.
I mean, it's redundant to speak about basis without saying that the base is numerable or countable, since by definition every topology of a topological space fulfills the properties you are describing in your definition.
To see that $S^{2}$ is second countable (have a basis), use the fact that $\mathbb{R}^{n}$ is second countable and try to use a basis to construct a new one in $S^{2}$. 
