Consider an infinite dimensional Banach space $X$. Let $B(0,r_0)$ be the ball with radius $r_0$. We know that the ball $B(0,r_0)$ is not relatively compact, so it is not totally bounded. This implies that there exists a radius $\rho>0$ such that any finite number of balls with radius $\rho$ cannot cover $B(0,r_0)$, of course we must have $\rho<r_0$, because $B(0,r_0)$ can cover itself.
My question is what can we say about numbers $r$ such that $\rho<r<r_0$. Can $B(0,r_0)$ be covered with a finite number of balls with radius $r$ ?
No, it cannot. For any $x_1, \ldots, x_n \in X$, the linear span $V$ of $x_1, \ldots, x_n$ is finite-dimensional, so it is not all of $X$, and by Hahn-Banach there is a nonzero continuous linear functional $f$ such that $f = 0$ on $V$. Take $y \in X$ with $0 < \|y\| < r_0$ so that $f(y) > r \|f\|$. Now $$\|y - x_j\| \ge \|f\|^{-1} |f(y - x_j)| = \|f\|^{-1} |f(y)| > r$$ Thus $y \in B(0,r_0)$, but it is not covered by the balls $B(x_i, r)$.