I'm studying Differential Manifolds using Manfredo do Carmo's Book (Riemannian Geometry) and although I see no mention of this in Do Carmo's book, it's really easy to see a Riemannian Manifold as a metric space (if $M$ is the manifold, using $d_M(x,y) = \inf\{L(\gamma) \colon \gamma$ is a curve on $M$ and $a,b \in \gamma([0,1])\}$, where $L(\gamma)$ is the length of the curve $\gamma$.
Is the topology induced in $M$ (viewed as a set) by the metric $d_M$ the same of $M$ as topological space?
My problem to solve this self-proposed question was analyzing the open balls of $(M,d_M)$. References and Hints are appreciated.
First, you need to stipulate that $M$ is connected for the distance function to turn $M$ into a metric space.
When $M$ is connected, the answer is yes, the metric topology is the same as the given manifold topology. The proof boils down to showing that the Riemannian metric is uniformly comparable to the Euclidean metric in small coordinate balls. You can read a proof in my Introduction to Smooth Manifolds (2nd ed., Theorem 13.29).