Compact subsets of Banach spaces Is a theorem of Mazur that a closed subset $K$ of a Banach space $X$ is compact if, and only if, there is a sequence $(x_n)$ of $X$ such that $|x_n|\to 0$ and $K\subseteq \overline{\operatorname{conv}}\{x_n:n\geq 0\}$, where $\overline{\operatorname{conv}}\{x_n:n\geq 0\}$ denotes the clousure of the converx hull of the sequence $(x_n)$. 
I'm trying to understand the proof given in the book Classical Banach Spaces I, by  J. Lindenstrauss and  L. Tzafriri. 
If $K$ is a compact subset of $X$, then there is points $x_{1,1}\dots x_{n_1,1}\in X$ such that 
$$2K\subseteq \bigcup_{i=1}^{n_1}B_{\frac{1}{4}}(x_{i,1})$$
Now, why must
$$K_2 = \bigcup_{i=1}^{n_1}\left[B_{\frac{1}{4}}(x_{i,1})\cap 2K-x_{i,1}\right]$$
be compact?
He then, by induction, obtain, for each $j\geq 2$, points $x_{1,j},\dots,x_{n_j,j}\in X$ such that 
$$K_j = \bigcup_{i=1}^{n_j}\left[B_{\frac{1}{4^{j}}}(x_{i,j})\cap 2K_{j-1}-x_{i,j}\right]$$
is compact.
Given a $x\in K$ there is a $1\leq i_1\leq n_1,\dots, 1\leq i_{k}\leq n_k$ such that 
$$x - \sum_{j=1}^{k}\frac{x_{i_j,j}}{2^j}\in \frac{1}{2^k}K_{k+1}$$
Why must $x\in \overline{\operatorname{conv}}\{x_{i,j}: j\geq 1\text{ and } 1\leq i\leq n_j\}$?
Also I would like to know if there is another proof of this fact.
 A: 
Now, why must
  $$K_2 = \bigcup_{i=1}^{n_1}\left[B_{\frac{1}{4}}(x_{i,1})\cap 2K-x_{i,1}\right]$$ be compact?

$2K$ is the image of the compact set $K$ under a continuous function and is therefore compact. $B_{\frac{1}{4}}(x_{i,1})\cap 2K$ is a closed subset of a compact set (I am assuming these are closed balls), so it is also compact. $B_{\frac{1}{4}}(x_{i,1})\cap 2K-x_{i,1}$ is again the image of the previous compact set under a translation, so it too is compact. Finally, the finite union of compact sets is compact. 

Given a $x\in K$ there is a $1\leq i_1\leq n_1,\dots, 1\leq i_{k}\leq n_k$ such that $$x - \sum_{j=1}^{k}\frac{x_{i_j,j}}{2^j}\in \frac{1}{2^k}K_{k+1}$$
  Why must $x\in \overline{\operatorname{conv}}\{x_{i,j}: j\geq 1\text{ and } 1\leq i\leq n_j\}$?

Because the radii of the balls in the definition of $K_k$ is $4^{-k}$, and each of these balls has been translated to the origin, so their union must also have radius $4^{-k}$. That is, $K_k \subseteq B_{\frac{1}{4^k}}(0)$. So as $k \to \infty$, the sums approach $x$. Since the sums are in $\operatorname{conv}\{x_{i,j}: j\geq 1\text{ and } 1\leq i\leq n_j\}$, $x$ must be in its closure.
Practically no result in mathematics has only one proof, but my own searching hasn't uncovered a different one.
