The diameter of S is the least upper bound of |x,y| for x and y in S. Note that the diameter of a circle or sphere agrees with its traditional definition. The diameter of an n dimensional unit cube is sqrt(n). The diameter of an equilateral triangle, one unit on a side, is 1, even though this triangle will not fit into a circle of diameter 1. A set with diameter d always fits in a ball of radius d.
A set is totally bounded if it has a finite cover of ε-balls for every ε > 0. A totally bounded set is always bounded. If the set is covered by k balls of radius 1, find the ball farthest from the origin; the far edge of that ball is the bound for the set.
Let's cover a region in Rn with balls of radius ε. Let t = ε/sqrt(n) and build a lattice, a regular grid, with spacing t. Let each of these lattice points be the center of an ε-ball, and the space is covered. A larger ball of radius r can be covered with finitely many ε-balls for any ε, hence every ball, and every bounded set, is totally bounded.
If we switch to Ej it's a different story. Try to cover the ball of radius 3, centered at the origin, with balls of radius 1. Build a sequence of points by setting each coordinate in turn to 2, while all the others are 0. All these points lie within the ball of radius 3, and each point is 2.818 distance from the others. Each must be covered by a separate ball of radius 1, requiring infinitely many balls. Within this metric space a bounded set need not be totally bounded.