Is it true that every totally geodesic surface $S$ of $\mathbb{R}^3$ is contained in a plane?
Locally, this is clear since geodesics in $S$ must be straight lines, and locally the exponential map is a diffeomorphism. However, I am not sure how to 'glue' all the different planes into a single global one?
Consider a geodesically convex open set. If it contained four non coplanar points, then the segments from one of them to the other three give you three geodesics through that point whose tangent vectors there are linearly independent, and this is impossible.
This means that every point in your surface has an open neighborhood contained in a plane, and therefore the open planar subsets of the surface cover it.
Now two planar subsets which intersect in at least three non collinear points are contained in the same plane, and clearly the intersection of two open planar sets of your surface satisfies that condition. Use that (and connectedness of the surface) to conclude.