In introduction to smooth manifolds by John M. Lee lemma 2.16 states that:
Every topological manifold admits a countable, locally finite cover by precompact open sets.
And a proof starts like this:
Let ${Ui}$ be any countable cover of M by precompact open sets (for example, small coordinate balls will do).
How can we take coordinate balls if we don't know whether charts map open subsets of manifold to open balls? We only know charts map open subsets of manifold to open subsets of $R^n$, not necessarily open balls.
I know that topological manifold has countable basis, but are these basis sets precompact?
Can you give me an example of manifold with atlas such that there is a chart or there are many charts that map open subsets of manifold to open subsets of $R^n$ but these open subsets of $R^n$ are not open balls?
Let us prove that each $x \in M$ has a precompact open neighborhood.
Let $B_r(y)$ denote the open ball in $\mathbb R^n$ having center $y$ and radius $r$.
Take any chart $\phi : U \to U'$ with $x \in U$. There exists $r > 0$ such that $B_r(\phi(x)) \subset U'$. Then $V' = B_{r/2}(\phi(x))$ is open neighborhood of $\phi(x)$ whose closure $C$ is copact and contained in $U'$. Then $V = \phi^{-1}(V')$ is an open neigborhood of $x \in M$ whose closure is $\phi^{-1}(C)$ which is compact (since $\phi$ is a homeomorphism).
What we have done here is to shrink the codomain $U'$ of a given chart $\phi$ to a small ball whose closure is compact and contained in $U'$.
Let us next prove that $M$ has a countable basis $\mathcal B'$ consisting of precompact open sets. This basis can be taken as the desired cover.
Let $\mathcal B$ be a countable basis for $M$. Define $$\mathcal B' = \{ B \in \mathcal B \mid \overline B \text{ is compact}\}. $$ This set is clearly countable. Let us show that it is basis which means that for each $x \in M$ and each open neighborhood $U$ of $x$ there exists $B \in \mathcal B'$ such that $x \in B \subset U$.
Take any precompact open neighborhood $U^*$ of $x$. There exists $B \in \mathcal B$ such that $x \in B \subset U \cap U^*$. Then $\overline B$ is a closed subset of $M$ which is contained in the closure of $U^*$ which is compact. Hence $\overline B$ is compact.