Is the metric on the circle, induced from the plane, not a flat one?

The Wikipedia page you have cited is wrong about metrics on the circle. The Wikipedia page on Curvature correctly points out that 1-dimensional curves do not have an intrinsic curvature.

It depends on which kind of "metric" you're looking at.

If we view the unit circle and the plane as Riemannian manifolds, then the Riemannian metric on the circle induced by its embedding in the plane is indeed flat. Here we're speaking about Riemannian metrics, which work only locally.

However, if you take those two Riemannian manifolds and derive a global distance function on each of them as the geodesic distance between two points, you make them into metric spaces. In the case of the plane, this produces the usual Euclidean distance.

However, as metric spaces, the metric on the unit circle is not the same as the metric it gets as a subset of $\mathbb R^2$.

Two opposite points on the unit circle has distance $\pi$ in the metric of geodesic distance within the circle, whereas their distance in the metric inherited from the metric space $\mathbb R^2$ is $2\neq \pi$.

I think you are correct but the subtle point here is the definition of curvature. You can define curvature of, e.g., a plane curve, as the rate of change of a unit normal vector field to the curve. Note that this quantity is extrinsic, i.e., it depends on a choice of embedding of the plane curve. For example, a circle of radius $r$ in $\mathbb{R}^2$ has extrinsic curvature $\frac{1}{r}$ in this sense

On the other hand, there is an intrinsic notion of curvature of a Riemannian manifold that can be computed purely from the metric and this is indeed $0$ for the circle independent of the choice of metric (the Riemann curvature tensor vanishes unless the dimension is $\geq 2$ but formal reasons). So, one would state that the curvature of a circle is $0$, if there is no choice of embedding in an ambient space apparent.

Hope this helps!