The Sphere Does Not Admit a Metric Isometry into a Euclidean Space.

Consider $X$, set of four points in $S^2(1)$ : $|pq|=|qr|=|pr| = \frac{\pi}{2}$ and $|qt|=|tr|=\frac{\pi}{4},\ |pt|= \frac{\pi}{2}$. So there is no isometric embedding $f$ from $X$ to $\mathbb{E}^3$.

[Add] In the above we have shown that there is no isometry from $T$ to Euclidean space where $T$ is equilateral triangle of side length $\frac{\pi}{2}$ in a sphere. Hence sphere is not flat.

In the paper, we can doubt that there may be local isometry. Hence it takes a small equilateral triangle $T'$ in a sphere. A little calculation (But idea is same to the above) show that there is no isometry around $T'$. Hence the paper concludes that sphere is not locally flat.

Take the equatorial circle $S^1\subseteq S^2$. If $S^2$ is embedded in the strongly isometric sense you described, then $S^1$ is also. But there is no strongly isometric embedding of a circle in Hilbert space since such an embedding would have to be a straight line (dealing with Gaussian curvature is a red herring).