I can't understand the end of the proof of the following lemma from "Introduction to smooth manifolds" by Lee
Lemma 1.10: Every topological manifold has a countable basis of precompact coordinate balls
In order to prove it he takes a manifold $M$ which is covered by a countable subcover $\{(U_i,\phi_i)\}$.
A the end of the proof he concludes that, given a precompact ball $V \subset U_i $, $\bar{V}$ is compact in $U_i$ and $\bar{V}$ is closed in $M$. And this is clear for me.
I don't understand the sentence that concludes the proof: " (given that $\bar{V}$ is closed in $M$) It follows that the closure of $V$ in $M$ is the same as its closure in $U_i$, so $V$ is precompact in $M$ as well."
What does it mean "is the same as"? How can I conclude that $V$ is precompact?
Can anyone help me?
It means the following:
Give $U_i$ the subspace topology. Then as $V \subseteq U_i$, we can take the closure of $V$ inside $U_i$ with the subspace topology: $cl_{U_i}(V)$. But also $V \subseteq M$, so we can take the closure of $V$ in $M$, which is $cl_M(V)$. The author claims that
$$cl_{U_i}(V) = cl_{M}(V)$$