Prove that the dimension of $X$ is finite 
Let $X$ be a normed space and suppose there are $n$ points $\lbrace x_{k} : 1\leqslant k\leqslant n\rbrace$
and a number $r>1$ such that
$$B(0,r) \subset \bigcup_{k=1}^n B(x_{k},1)$$
Prove that the dimension of $X$ is finite.

I have tried using Riesz's theorem, we know that if the closed unit ball is compact by this theorem we have that the dimension of $X$ is finite, but I don't know how to prove the compacity of the closed unit ball.
 A: Let $$r'=\frac{1+r}{2}$$
so that $1 < r' < r$, and let $0 < \varepsilon < r'-1$.
Let $V = \mathrm{Vect}(x_1,..., x_n)$, and suppose that $X \neq V$.
Then there exists $y \in X \setminus V$. Let $d=d(y,V)$ the distance from $y$ to the subspace $V$ : because $V$ is finite-dimensional, hence it is closed, hence $d>0$. By definition of $d$, there exists $z \in V$ such that
$$d \leq ||y-z|| \leq (1+\varepsilon)d$$
Let $$z_0 = r'\frac{y-z}{||y-z||}$$
Now, for all $x \in V$, one has
$$\left|\left| x-z_0\right|\right| = \left|\left| x-r'\frac{y-z}{||y-z||}\right|\right|   = r'\frac{\left|\left|(x\frac{||y-z||}{r'}+z)-y\right|\right|}{||y-z||}$$
But $$x\frac{||y-z||}{r'}+z \in V$$ so the numerator is greater than $d$ ; and the denominator is less than $(1+\varepsilon)d$ by definition : you deduce $$\left|\left| x-z_0\right|\right| \geq \frac{r'}{1+\varepsilon}$$
In particular, $$d(z_0, V) \geq \frac{r'}{1+\varepsilon} > 1$$ by definition of $\varepsilon$. But by hypothesis, because $||z_0||=r' < r$, then $z_0 \in B(0,r)$, so there exists $i \in \lbrace 1, ..., n \rbrace$ such that $z_0 \in B(x_i, 1)$, i.e. such that $d(z_0, x_i) < 1$. So one must also have $d(z_0, V) < 1$. Contradiction.
So the hypothesis $X \neq V$ is not true, so $X=V$ is a finite-dimensional vector space.
