@PeterTamaroff So let us just see how simple this problem can be:
We want to show that inside every nbhd, there is a closed nbhd of a point $a\in X.$ We draw a picture. We know we can draw an open ball around $a$ and a good candidate to be a closed nbhd would be the closed ball with half the radius, but we need to show that is indeed a subset. So we need to show $d(a,x)> r/2$ is impossible for points in the closure of $B(a,r/2).$ *Back to this in a second. With that, $d(a,x) \leq r/2 < r$ so $x\in B(a,r)$ as required.