This is part of exercise 9.2, from Bachman's `Functional Analysis'. Doesn't this just follow from the definition of a norm (i.e. it satisfies the triangle inequality)? Seems a bit funny to me that he'd set such a trivial exercise, so I feel like I must be missing something. Furthermore, I haven't used completeness of the space in any way whatsoever, and I feel that if he's specified it's a Banach Space, then I should have.
Why doesn't that inequality follow trivially from the definition of a norm?
The first lesson you need to learn in functional analysis is that infinity is tricky. Arguably, most of the tough questions in functional analysis are "can we still get away with this trick in the infinite case?"
Explicitly, the triangle inequality tells you that $\|x + y\| \leq \|x\| + \|y\|$. By induction, we can deduce that in any finite sum, we have $$ \left \|\sum_{n=1}^N x_n\right\| \leq \sum_{n=1}^N \|x_n\| $$ However, we can not simply state that this is true when "$N = \infty$". Presumably, he has told you as an assumption to the problem that $\sum_{n=1}^\infty \|x_n\|$ is finite. With this, we may use the completeness of a Banach space to conclude that the sum $x := \sum_{n=1}^\infty x_n$ is defined as an element in the Banach space, and that $\lim_{N \to \infty}\sum_{n=1}^N x_n = x$.
Armed with that knowledge, we can now use the continuity of the norm to conclude that $$ \left \|\sum_{n=1}^\infty x_n\right\| = \left \|\lim_{n \to \infty}\sum_{n=1}^N x_n\right\| = \lim_{n \to \infty} \left \|\sum_{n=1}^N x_n\right\| \leq \lim_{n \to \infty}\sum_{n=1}^N \|x_n\| = \sum_{n=1}^\infty \|x_n\| $$ as desired.