$S=\{x\in X: \|x\|=1\}$ cannot be covered by a finite family of closed balls in X I am working with an exercise and have som questions.
The exercise is
Let X be an infinite dimensional normed vector space  over $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$
Show that one cannot cover the unit sphere $S=\{x\in X: \|x\|=1\}$ with a finite family of closed balls in $X$ such that none of the balls contains 0.
I am thinking to show that S is not covered by a finite family of closed balls in X such that none of the balls contains 0. To show this I am thinking that I have to show $S\not\subset \cup \{x\in X| d(x,p)\leq r\}$ and to show this I am thinking to take a $x\in S$ and show that $x \notin \cup \{x\in X| d(x,p)\leq r\}$.
I note that for $x\in S$ I have that $\|x\|=1$. And to show that $x\notin \cup \{x\in X| d(x,p)\leq r\}$ I am thinking that I have to show that $d(x,p)>r$ . And to show this I am thinking that I somehow have to use Riesz lemma, but to be able to use this lemma I have to have that $\cup \{x\in X| d(x,p)\leq r\}$ is a closed proper subspace of X. It is obviously closed but I don't know how to show that it is a proper subspace of X.
But after I have argued that it is a proper subspace of X then I am thinking that I have from Riesz lemma that there exist a $x\in X$ with $\|x\|=1$ such that $\|x-y\|\geq a$ for all $y\in 
\cup \{x\in X| d(x,p)\leq r\}$, where $0<a<1$. Since we have that $p\in \cup \{x\in X| d(x,p)\leq r\}$ then we have that $\|x-p\| \geq a$, and what we want to show is that $d(x,p)>r$ i.e. $\|x-p\|>r$. So dont now how I get from $\|x-p\| \geq a$ to $\|x-p\|>r$
 A: I found another, easier proof using the Hahn-Banach theorem that works in the general case. Let $B_i$ for $i=1,\ldots,n$ be closed balls not containing $0$. These balls are closed convex sets. Hence, we can find continous functionals $\lambda_i$, s.t. $\mathrm{Re}\,\lambda_i(x) \geq 1$ for $x \in B_i$. The vector space $V = \cap_{i=1}^n \mathrm{ker}(\lambda_i)$ does not intersect any of the $B_i$. But $V \neq 0$, because $X$ is infinite-dimensional. So we find an $x\in V \cap S$. Therefore, no finite number of closed balls not containing $0$ can cover $S$.
A: In the union $\cup\{x \in X, d(x,p) \leq r \}$, what are the values of $p$? I assume that $p$ and $r$ run over a finite set. Let me denote the closed ball around $p$ with radius $r$ by $B(p,r) = \{x \in X, d(x,p) \leq r \}$ and assume by contradiction that $S \subset \cup_{i=1}^n B(p_i,r_i)$. Since the balls are centered on $S$ and cannot contain $0$, we have $||p_i|| = 1$ and $0 < r_i < 1$.
Now, let $e_1$ be a unit vector. Then, $e_1 \in B(p_{i_1},r_{i_1})$ for some $1 \leq i_1  \leq n$. Riesz's lemma now finds an $e_2 \in S$, s.t. for $y \in lin\{e_1,p_{i_1}\} = W_1$ we have $||y-e_2|| \geq r_{i_1}$. In particular, $e_2 \notin B(p_{i_1},r_{i_1})$.
Iterating this construction, we find closed subspaces $W_k = lin\{e_1,p_{i_1},\ldots,e_k,p_{i_k}\}$, $e_{k+1} \notin W_k \cup \cup_{j=1}^k B(p_{i_j},r_{i_j})$ and $i_r \neq i_s$ for $r \neq s$. Since $X$ is infinite dimensional, $W_k$ are always proper subspaces of $X$. But then we obtain a contraction when $k=n$.
