If you have a non-zero normed vector space over the field $k = \mathbf{R}$ or $\mathbf{C}$ then you always have a unit vector $u$ (that is, a vector of norm $1$) and then vectors $ru$ and $-ru$ have distance $2r$, so that you have equality indeed. On the zero vector space, the equality is false though, as soon as $r>0$.
But beware, this is false in general. Consider the field $k = \mathbf{Q}_p$ of $p$-adic numbers, for a fixed prime number $p$, and let simply $V = k$ endowed with the $p$-adic absolute value $|\cdot|_p$. Consider then $B_F (0, 1)$ for instance, and let $x,y$ be in it. As $|\cdot|_p$ is non-archimedean, you have $d(x,y) = \|x-y\| = |x-y|_p \leq \max(|x|_p,|y|_p)\leq 1$, and the diameter will never reach $2$.
Let $D(x,r) = \{ y | \|x-y\| \le r \}$ and $B(x,r) = \{ y | \|x-y\| < r \}$.
We have $B(x,r) \subset D(x,r)$ for all $r$.
Since the norm is continuous we see that $D(x,r)$ is closed and hence $\overline{B}(x,r) \subset D(x,r)$.
Now suppose $y \notin \overline{B}(x,r)$. Since $\overline{B}(x,r)$ is closed
there is some $\epsilon>0$ such that $B(y,\epsilon)$ does not intersect $\overline{B}(x,r)$ (and hence does not intersect $B(x,r)$). It follows that
$\|x-y\| \ge r + \epsilon$ and
so $y \notin D(x,r)$. Hence $D(x,r) \subset \overline{B}(x,r)$.
Addendum: To show why $\|x-y\| \ge r + \epsilon$:
Let $\phi(t) = t y +(1-t)x$. For $0 \le t_1 \le t_2 \le 1$ we have $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$.
Note that $\phi(t) \in B(x,r)$ if $t \in [0,{r \over \|x-y\|})$ and
$\phi(t) \in B(y,\epsilon)$ if $t \in (1-{\epsilon \over \|x-y\|}), 1]$, and we must have $t_1={r \over \|x-y\|} \le t_2=1-{\epsilon \over \|x-y\|}$ since the balls do not overlap.
Finally $\|x-y\|= \|\phi(0)-\phi(1)\| \ge \|\phi(0)-\phi(t_1)\| + \|\phi(t_2)-\phi(1)\| = r+ \epsilon$.
Another take:
Consider $\phi(t)$ for $t \in [0,1]$. For the $t \in [0,{r \over \|x-y\|})$ part we have $\phi(t) \in B(x,r)$ and for $t \in (1-{\epsilon \over \|x-y\|}), 1]$ we have $\phi(t) \in B(y,\epsilon)$. The balls do not overlap, so the intervals $[0,{r \over \|x-y\|}), (1-{\epsilon \over \|x-y\|}), 1]$ are disjoint and the combined length of the intervals is ${r+ \epsilon \over \|x-y\|}$.
Since $\|\phi(t_2)-\phi(t_1)\| = (t_2-t_1) \|x-y\|$ for $0 \le t_1\le t_2 \le 1$ we see that
$\|x-y\|=\|\phi(1)-\phi(0)\| = (1-t_2)+(t_2-t_1)+(t_1-0)) \|x-y\| \ge (1-t_2)+(t_1-0) \|x-y\|$, and since $(1-t_2)+(t_1-0) = {r+ \epsilon \over \|x-y\|}$ we have the desired result.
Best Answer
For any $x,y \in B_\epsilon (a)$, $||x - y|| \leq \rm{diam}\, B_\epsilon (a)$. So, $k = ||x - y|| \leq \rm{diam}\, B_\epsilon(a) < k$, a contradiction.