How do I deal with the basis in the proof of Lemma 1.10 in Introduction to smooth manifolds by John Lee? This is Lemma 1.10 in Introduction to smooth manifolds by John Lee, 2nd edition.

Lemma 1.10. Every topological manifold has a countable basis of precompact coordinate balls.
Proof. Let $M$ be a topological $n$-manifold. First we consider the special case in which $M$ can be covered by a single chart.
Suppose $\varphi: M \to \hat U \subseteq \mathbb{R}^n$ is a global
coordinate map, and let $\mathscr{B}$ be the collection of all open
balls $B_r(x)\subseteq  \mathbb{R}^n$ such that $r$ is rational, $x$
has rational coordinates, and $B_{r'}(x)\subseteq  \hat U$ for some
$r' > r$. Each such ball is precompact in $\hat U$, and it is easy
to check that $\mathscr{B}$ is a countable basis for the topology of
$\hat U$. Because $\varphi$ is a homeomorphism, it follows that
the collection of sets of the form $ \varphi^{-1}(B)$ for$ B \in \mathscr{B}$ is a countable basis for the topology of $M$;
consisting of precompact coordinate balls, with the restrictions of
$\varphi$ as coordinate maps.

My question is about checking  that the $ \varphi^{-1}(B)$ for$ B \in \mathscr{B}$ are a countable basis for the topology of $M$

*

*First of all is it correct that to prove that a set is a basis of a topology it suffices to prove that the any element of the topology is a union of element of basis elements?
According to wikipedia the following is the definition.


But I think I have read proofs were they just do as I described. Are they equivalent definitions?


*About the question

I guess I can prove it like this:
Let $\mathscr{B}$ be a basis for the topology of $\hat U$. Let $U \in \tau_M$, then $ \varphi(U) \in \tau_{\hat U}$, because $\varphi$ is a homeomorphism  and $\varphi(U)=\bigcup_{i \in I}B_i,  B_i \in \mathscr{B}  \implies U = \varphi^{-1}(\bigcup_{i \in I}B_i)=\bigcup_{i \in I}\varphi^{-1}(B_i) $ and since I have written any open set of $\tau_M$ as union of elements of $\{\varphi^{-1}(B):, B \in  \mathscr{B} \}$, this last set is a basis for $\tau_M$. Is this correct?
Thank you!
 A: If the basis elements are in the topology, and every open set is a union of basis elements, then the topology is exactly the topology generated by the basis elements.
This is what it means to be a basis for the topology, I can express every open set as a union of the sets in my basis, and every union of basis elements is open in the topology. This is the definition of a basis for a topology.
A lot of heavy lifting is done by the requirement that the basic open sets be open, since the singletons are not a basis for every topology.
Some more intuition: I can make a topology out of any collection of subsets I like in two equivalent ways. I can either consider the intersection of all topologies in which my subsets are contained, or I can add in the whole space to my collection, and take, as the topology, the set of unions of finite intersections of the subsets in my collection.
Such a set is a basis for the topology that it generates only if the sets cover my space (B1), and the intersection of any two sets in my base is a union of elements in my base (B2, check this). Then every element of the topology they generate (in one of the two equivalent senses above) is exactly a union of the elements in the base, rather than having to potentially intersect them with each other and the whole space first.
Your proof is otherwise correct.
