Let $(X, \|.\|)$ be an NLS, $x\in X$ and $0 < rIn my text I've found the problem:

Let $(X, \|.\|)$ be an normed linear space, $x\in X$ and $0 < r<s.$ Show that $B(x, r)\subsetneq B (x, s).$

I can see if $\exists~y\in X-\{0\}$ then $x+\frac{r}{\|y\|}y\in B(x,s)-B(x,r).$
But what if no such nonzero $y$ exists ? i.e. what if $X$ is the trivial vector space over $\mathbb R$ or $\mathbb C?$
 A: I think the standard definition of normed linear space allows $X = \{0\}$, since this case might come up in applications, viz. as the kernel of some injective operator or the image of 
an operator which has zero as an eigenvalue.  So I think we need to assume $X \ne \{0\}$, or
(equivalently) that $X$ contains a non-zero element.  (Incidentally, the widipedia definition of NLS allows $X = \{0\}$, see http://en.wikipedia.org/wiki/Normed_vector_space.)
Assuming this to be the case, the OP's argument appears to be basically correct.  Just for the record, here's my way of getting at it:  pick any $y \in X$, $y \ne 0$, so that $||y|| \ne 0$  Pick $\alpha$ real, $0 < \alpha <\frac{r}{||y||}$; then $||\alpha y|| = \alpha ||y|| < r$, so $\alpha y \in B(0, r)$.  Pick real $\rho$ with $r < \rho < s$.  Set $\beta = \frac{\rho}{||\alpha y||}$.  Then $||\beta \alpha y|| = \beta ||\alpha y|| = \rho$, so $\beta \alpha y \in B(0, s)$; since $\rho > r$, $||\beta \alpha y|| \notin B(0, r)$.  Now just translate everything by $x$.  Incidentally, this "proof" works whether the balls $B(x, d)$ are open or closed.
