The riemannian distance topology is the manifold topology

Choose a coordinate ball $B$ of radius $r$ centred at $p$ whose closed ball lies in a chart. The compact comparison gives constants $c,C>0$. As in the metric theorem’s first-exit calculation, any path from $p$ to a point outside $B$ has length at least $\sqrt{c}r$: restrict to its first exit and integrate the Euclidean displacement bound. Thus $B_{d_g}(p,\sqrt{c}r)\subseteq B$. Choosing the closed coordinate ball inside any given manifold neighbourhood proves that neighbourhood contains a metric neighbourhood.

Conversely, for $q$ in that convex coordinate ball the coordinate straight segment has length at most $\sqrt{C}\mathrm{\mid }x(q)-x(p)\mathrm{\mid }$. Hence $d_g(p,q)\le \sqrt{C}\mathrm{\mid }x(q)-x(p)\mathrm{\mid }$, so the coordinate ball of radius $\min (r\mathrm{/}2,\epsilon \mathrm{/}(2\sqrt{C}))$ is inside $B_{d_g}(p,\epsilon )$. This proves the opposite neighbourhood inclusion. Half-balls are convex and give the same estimates at a boundary; a point or empty manifold has the unique topology.