Well it is actually intuitively evident I guess. Consider a function $ f$ that gives the distance between two points lying on the sphere. (automatically consider a mapping from $ f$ to $ g$ where $ g$ gives the length of the curve between the two points). So, $ AB \le OA + OB$ ($ A,B$ are points on the sphere) and $ O$ is the center of the sphere. $ AB \le 2$. But then you can say that if you want $ f$ to attain its maximum, then automatically your $ A,B$ will tend to be diametrically opposite by the equality of the skew triangle's lengths or whatever (I dont know how it is called). But then, we are given that $ AB<2$, so we can say that for equality itself, the points are diametrically opposite at the maximum, so $ f$ will make sure that $ A,B$ lie entirely within one hemisphere.So we are done.

Agr_94_Math wrote: Well it is actually intuitively evident I guess. Consider a function $ f$ that gives the distance between two points lying on the sphere. (automatically consider a mapping from $ f$ to $ g$ where $ g$ gives the length of the curve between the two points). So, $ AB \le OA + OB$ ($ A,B$ are points on the sphere) and $ O$ is the center of the sphere. $ AB \le 2$. But then you can say that if you want $ f$ to attain its maximum, then automatically your $ A,B$ will tend to be diametrically opposite by the equality of the skew triangle's lengths or whatever (I dont know how it is called). But then, we are given that $ AB < 2$, so we can say that for equality itself, the points are diametrically opposite at the maximum, so $ f$ will make sure that $ A,B$ lie entirely within one hemisphere.So we are done. That's not rigorous at all. Why would you want $ f$ to attain its maximum?

Well it is closed continuous and bounded region over which $ f$ is defined. So automatically it attains its global maximum due to Weirstrass theorem.

Well, let's consider the most disadvantageous case, when the curve is an arc of great circle. i.e. belonging to the intersection of the sphere with a plane passing through its center, let $A, B$ be its extremities and draw an equatorial plane containing $A$ and being perpendicular onto the plane OAB. One of the two hemispheres thus constructed contains $AB$ entirely, since the length of the arc $AB$ is lesser that a semicircle. Best regards, sunken rock

I think you are all misunderstanding it as a simpler problem than it is. It is saying ANY curve of length <2 is contained in a hemisphere, but this is true because a hemisphere can contain anything that has maximum length 2