Well, I'm a bit confused: If you have a regular surface M at $\mathbb {R}^3$ you have, of course, the Euclidean metric induced. On the other hand, if you have a local parametrization $ \phi: U \to V $ with U open in the plane and V in M, you can induce a metric in M by the Euclidean metric in $ \mathbb {R}^2$. Namely
$$g_{p}(u, v) = <(d\phi^{-1}_{p}.u, d\phi^{-1}_{p}.v>$$
I think that's right. But in one exercise, the assertion is that: "No open set U $\subset S^{2}$ is isometric to an open set W $\subset \mathbb{R}^{2}$ with the Euclidean metric. " I think the equation above defines a local isometry and agrees with the statement because not necessarily $ V $ is open in $ M $. Right?
Thanks in advance