I am trying to understand the proof of
Let M be a smooth manifold. Every open cover of M has a regular refinement.
The proof begins as follows [Lee] :
Let $X$ be an open cover of $M$, and let $\{V_j\}$ be a countable, locally finite cover of M by precompact open sets. For each $p \in M$ let $(W_p,\phi_p))$ be a coordinate chart centred at p such that:
1) $\phi_p(W_p)=B_3(0)$
2) $W_p$ is contained in one of the open sets of $X$ and
3) if $p$ in $V_j$,then $W_p \subset V_j$ as well. (The last condition is possible because of the local finiteness of $\{V_j\}$)
...
My Problem:
As I have little knowledge of topology, In my mind the openess of $V_j$ suffices to make the last condition possible and I do not understand why the local finiteness is necessary.
3) means that you can choose each $W_p$ in a uniform way with respect to the cover. I.e. as $p$ is only contained in finitely many $V_j$, there is a nice ball around $p$ contained in all $V_j$, i.e. in the intersection.
Another way to say this comes basically immediately from the definition of topology: Finite intersections of open sets are open, however infinite intersections don't have to be.
An easy example would be: $\{ (-\frac 1n , \frac 1n),n\in \mathbb N\}$ forms an open cover of $(-1,1)$ however $0$ does not have an open neighborhood which is contained in all open sets, that is to say $$\bigcap_{n\in \mathbb N} (-\frac 1n, \frac 1n)=\{0\}.$$