Riemannian manifold as a metric space I am looking for a reference to the following simple statement; it must be classical. (It is easy to proof, but I want to have a reference.)

A metric space $X$ that corresponds to a Riemannian manifold $(M,g)$ completely determines the underlying smooth manifold $M$ and the metric tensor $g$.

 A: Isn't this the Myers-Steenrod theorem? "If $(M,g)$ and $(N,h)$ are connected Riemannian manifolds and $f:(M,d_g)\to(N,d_h)$ is an isometry, then $f:(M,g)\to(N,h)$ is a smooth isometry"
A: It was proven by Dick Palais. 

MR0088000 (19,451a) Reviewed Palais, Richard S. On the
  differentiability of isometries. Proc. Amer. Math. Soc. 8 (1957),
  805–807.
  53.2X

MathSciNet

@article {MR88000,
      AUTHOR = {Palais, Richard S.},
       TITLE = {On the differentiability of isometries},    JOURNAL = {Proc. Amer. Math. Soc.},   FJOURNAL = {Proceedings of the American
  Mathematical Society},
      VOLUME = {8},
        YEAR = {1957},
       PAGES = {805--807},
        ISSN = {0002-9939},    MRCLASS = {53.2X},   MRNUMBER = {88000}, MRREVIEWER = {K. Krickeberg},
         DOI = {10.2307/2033302},
         URL = {https://doi-org.ucc.idm.oclc.org/10.2307/2033302}, }

According to Palais, if I read his paper correctly, Myers and Steenrod proved the differentiability of isometries but Palais obtained an explicit description of smooth functions on the manifold from the metric geometry.
