If a (distance) metric on a connected Riemannian manifold locally agrees with the Riemannian metric, is it equal to the induced metric?

The answer is no. Take any non-convex region in the plane, and let the Riemannian metric be the ordinary Euclidean metric $ds^2=dx^2+dy^2$. Then define the new metric as the infimum of Euclidean diameters of curves connecting $a$ to $b$. This new metric coincides with the Riemannian metric locally, but does not coincide globally. This metric even has a name: Mazurkiewicz metric.

One can also construct a compact example by taking $M$ to be a ramified covering of the sphere, and pullback of the spherical metric. This Riemannian metric in general does not coincide with the Mazurkiewicz metric corresponding to it. (Diameter used in the definition is the diameter of the projection of the curve on the sphere. Projection of the Riemannian-shortest path on $M$ can have small diameter but large length).

In general, any distance $d$ which locally coincides with the Riemannian distance $d_R$ must satisfy $d\leq d_R$. Just break the curve on which the Riemannian distance is (almost) achieved into small pieces, and use the triangle inequality for $d$.

Take $d(x,y) = {\rm min}(|x-y|, 1)$ on $\mathbb{R}$,