Proof about cluster point a)Let $E$ be a subset of $R^n$. A point $a∈R^n$ is called a cluster point of E if $E∩B_r(a)$ contains infinitely many points for every $r>0$. Prove that a is a cluster point of $E$ if and only if for each $r>0$, $E∩B_r(a)$\{$a$} is nonempty.
b) Prove that every bounded infinite subset of $R^n$ has at least one cluster point.
 A: a.
If there exists a radius $r>0$ such that $E\cap \big(B_r(a)\setminus\{a\}\big)=\emptyset$, then $E\cap B_r(a)\subseteq\{a\}$, and $a$ is trivially not a cluster point.
Let us now assume that for every $r>0$ $E\cap \big(B_r(a)\setminus\{a\}\big)\neq\emptyset$. Then for $r_1=1$ we can pick a point $a_1\in E\cap\big(B_{r_1}(a)\setminus\{a\}\big)$.
Now, taking $r_2=\frac{|a_1-a|}{2}>0$ we can pick another point $a_2\in E\cap\big(B_{r_2}(a)\setminus\{a\}\big)$. Note that $r_2<r_1$ and $a_1\notin B_{r_2}(a)$ and in particular $a_2\neq a_1$.
Now we can consider $r_3=\frac{|a_2-a|}{3}$ and pick another $a_3\in E\cap\big( B_{r_3}(a)\setminus\{a\}\big)$ different from the previous two, and so on with $r_4=\frac{|a_3-a|}{4}$, etc.
Note that $r_n\to0$ as $n\to\infty$.
If $r>0$, then it suffices to take $N$ large enough so that $r_N<r$, so that, by construction of the sequence $\{a_n\}_{n\in\mathbb N}$, you have that
$$
\{a_n\}_{n\geq N} \subseteq B_r(a) \cap E
$$
and since $a_n$ are all distinct, then you found an infinite number of points in $B_r(a)\cap E$.
b.
If $A$ is an infinite bounded subset of $\mathbb R^n$, then its closure $\overline A$ is compact. Once you pick a sequence of distinct points $\{a_n\}_{n\in\mathbb N}\subseteq A$, you may extract a subsequence $\{a_{n_k}\}_{k\in\mathbb N}\subseteq\{a_n\}_{n\in\mathbb N}$ which converges to a point $a\in\overline A$. It follows from point a. and from the definition of limit that $a$ is a cluster point.
A: For part (a), if $\exists r'>0 \quad E\cap B_{r'}(a)/\{a\} = \phi $, then for $r=r'/2$, the set is empty and so a is not a cluster point. 
The proof of the other way is there in Rudin's Principles of Mathematical Analysis. Check out every open neighborhood has infinitely many points. The proof is by contradiction. Its actually quite direct.
For part (b), check out Weierstrass Bolzano Theorem.
