I apologize if the question may be trivial, but it is a fact that my textbook does not even mention and I, studying as a self-taught, do not have so many certainties.I believe the totally boundedness $ \implies $ boundedness implication is true in any metric space. I think I managed to prove it this way:
$ A $ totally bounded $ \implies $ there is a finite $ \epsilon $-grid for every $ \epsilon> 0 $. Choose one: $ E = \{g_1, g_2, ..., g_n \} \subset A \implies $ the distance between any element $ a \in A $ and a $ g_i \in E $ is $ d (a , g_i) \leq d (a, g_k) + d (g_k, g_i) <\epsilon + \max_{j = 1, ..., n} d (g_k, g_j) \implies A $ bounded because entirely contained in the ball $ B (g_i, \epsilon + \max_{j = 1, ..., n} d (g_k, g_j)) $.
Did I do something wrong?