Coordinate balls form a basis of a topological manifold

By [F1] choose a chart $(U,\phi )$ of $M$ at $p$. Since $O$ is open and $p\in U\cap O$, replacing $U$ by $U\cap O$ and $\phi$ by its restriction still gives a chart at $p$ whose image is the open set $\phi (U\cap O)$ in ${\mathbb{R}}^n$. So we may assume from the start that $U\subseteq O$.

Put $c:=\phi (p)$. Because $\phi (U)$ is open, [A1] gives $r>0$ with $\overline{B(c,r)}\subseteq \phi (U)$. Then $p\in {\phi }^{-1}[B(c,r)]\subseteq U\subseteq O$. This gives the required coordinate ball inside $O$.

The closure of ${\phi }^{-1}[B(c,r)]$ in $M$ is contained in ${\phi }^{-1}[\overline{B(c,r)}]$, and [A2] identifies ${\phi }^{-1}[\overline{B(c,r)}]$ with the homeomorphic image of the Euclidean closed ball $\overline{B(c,r)}$. For $n\ge 1$ this set is compact by [L1], hence its homeomorphic image is compact by [A2] and the smaller closure in $M$ is compact as a closed subset of a compact set. When $n=0$, the chart image is the one-point space ${\mathbb{R}}^0$, so the same conclusion is immediate.

Since step 2.1 works for every point $p$ and every open neighbourhood $O$ of $p$, the coordinate balls form a basis of the topology of $M$.