I am trying to show that every manifold (a Hausdorff and second countable topological space) is locally compact (has a precompact basis).
Consider an open set of a manifold, $M$. I must show that for each $x\in U$ we can find an open set in $M$ such that its closure is compact.
So let $x\in U$ where $U$ is an open set in $M$. Because $M$ is a manifold, we can find an open set contained in $U$ containing $x$ homeomorphic to an open Euclidean ball. Call this set $V$. Now since $V$ is homeomorpic to a Euclidean ball, we can find another open set, $G$ contained in $V$, containing $x$ such that the closure of $G$ in $V$ is compact.
To summarise, we have an open set $V$ in $M$ containing an open set $G$ and $\text{cl}_V{(G)}\subseteq V$ is compact. To get the result I need, I want to say that this mean the closure of $G$ in $M$ is compact. However I get stuck here.
My question is: how can I show that the closure of $G$ in $M$ is also compact? Am I in the right direction?
EDIT: I was thinking of arguing that the closure of $G$ in $M$ is contained in $V$ and then the result follows immediately since then the two closures are equal. Is this a true statement?