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.
The first part is fine. I’d give a little more detail for the second part:
Suppose that $X$ is sequentially compact, and let $S$ be an infinite subset of $X$. Then we may choose a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ of distinct points of $S$. Let $\tau=\langle x_{n_k}:k\in\Bbb N\rangle$ be a subsequence of $\sigma$ converging to $x\in X$. Then every nbhd of $x$ contains infinitely many terms of $\tau$ and hence infinitely many distinct points of $S$, so $x$ is a cluster point of $S$.
Best Answer
Hint Just one direction needs a proof
$\Leftarrow$ by contraposition: suppose that $E\cap B_r(a)\setminus\{a\}$ contains finitely many points $x_1,\ldots,x_s$ for some $r>0$ and let $d=\min_{i}d(a,x_i)$. Prove that $E\cap B_{d/2}(a)\setminus\{a\}$ is the empty set.