Check proof that every manifold has a basis of precompact open sets. I've written a prove of the theorem. I need someone to check if is correct and explain why Claim 1,3,7 are true.
Definition: Let $M$ be a topological manifold, Then the subset $U \subset M$ : $f(U)=B_r(x)=\{x:\mathbb{R^n} : \|x\|<r\}$ is called coordinate ball.

Definition: Let $(M,T)$ be a topological space, then the subset $U \subset M$ is called precompact if $ \overline{U}$ is compact.


Theorem: Every manifold $M$ has a basis of precompact coordinate balls.

Proof: We know around every $x \in M$ there exists an open neighborhood of $x$ : $U$ and a homeomorphism $f:U \rightarrow U' \subset \mathbb{R^n}$

Claim 1 Since $M$ has a countable basis (from the definition) then there are countable $(U_i)_{i=1}^{\infty}$ that cover $M$

Claim 2: There are countable charts $\{U_i,f_i\}_{i=1}^{\infty}$ s.t. $f_i : U_i \rightarrow U_i'\subset \mathbb{R^n}$

Now, we define the set:
$$ \mathcal{B}_i=\{B_r(x): (x \in U_i'\cap \mathbb{Q}^n \quad \text{and} \quad r \in \mathbb{Q}_+) \wedge( \forall r \in \mathbb{Q}_+ , \exists r'>r : B_{r'}(x) \subset U_i')\}$$
The second condition of the set $\mathcal{B}$ exist to guarantee that $\overline{B_r(x)_{\mathbb{R}^2}} =\overline{B_r(x)_{U_i'}}$

Now, for the set $\mathcal{B}_i$ we obtain that:
Claim 3: Consists of precompact sets.
Claim 4: It is countable, ($x,r$ are rational)

Also

Claim 5: Since $f_i$ are homeomorphisms, the sets $\{f_{i}^{-1}(B_r(x)\}$ have the same topological properties and

Claim 6: the set $\{f_{i}^{-1}(B_r(x)\}_{B_r(x) \in \mathcal{B}_i}$ cover $U_i \subset M$
So, finally we conclude that:

Claim 7: Every open set of $M$ can be expressed as a union of elements in $\mathcal{B}_i$, So
$$\{f_{i}^{-1}(B_r(x)\}_{B_r(x) \in \mathcal{B}_i}$$
Is a basis of $M$ whose elements are coordinate balls.
 A: Yes, your proof is correct.
In order to make this clear : you start by considering $\{(U_{\alpha},f_\alpha)\}_{\alpha\in\mathcal{A}}$ the collection of all coordinate charts in $M$, where $U_\alpha\subset{M}$ open, and $f_\alpha : U_\alpha\longrightarrow{f(U_\alpha)}={U'_\alpha}\subset\mathbb{R}^n$ homeomorphism.
$\textbf{Definition.}$ Let $X$ be a topological space. X is a Lindelöf space if each $\{U_\alpha\}_{\alpha\in\mathcal{A}}$ open cover of X admits $\{U_i\}_{i\in\mathbb{N}}\subset\{U_\alpha\}_{\alpha\in\mathcal{A}}$ a countable subcover.
$\textbf{Theorem.}$(Lindelöf) Let X be a second-countable space $\implies$ X is  Lindelöf space.
Applying the previous theorem, you obtain (Claim 1) a countable cover $\{U_i\}_{i\in\mathbb{N}}$ of $M$ consisting of coordinate charts, with $f_i : U_i \longrightarrow{f(U_i)}={U'_i}\subset\mathbb{R}^n$ homeomorphism.
You go on by defining $\mathcal{B}_i$, consisting of precompact sets because $\overline{B_r(x)}$ is a closed and bounded subset of $\mathbb{R}^n$, then is compact for Heine-Borel (Claim 3).
Finally, $\mathcal{B}_i$ covers $U'_i$, and the non empty intersection of two balls $B_r(x)$,$B_s(y)\in\mathcal{B}_i$ is a union of balls of the same type because you can always choose a rational $t<min\{r,s\}$ such that $B_t(z)\subset{B_r(x)}\cap{B_s(y)}$, for some $z\space{\in}\space{B_r(x)}\cap{B_s(y)}$. (This is a common argument in general topology).
Thus $\mathcal{B}_i$ is a base for $U'_i$, and so on $\{f^{-1}_i{B_i(x)}\}_{B_i(x)\in\mathcal{B_i}}$ is a base for $U_i$. (Claim 7) By taking the countable union of all these basis, you get a countable base of precompact coordinate balls for M.
If you want more details you can see Lemma 1.10 in "Introduction to Smooth Manifolds", Lee.
