Cluster points infinitely many points 
A point $x$ is a cluster point of a set $A\subset \mathbb{R}^p$ if and
  only if every neighborhood of $x$ contains infintely many points of
  $A$.

I have learned that by definition, a cluster point of a set $S\subseteq X$ is a point $x\in X$ such that for every neighbourhood $U$ of $x$, there is a point $y\in S\cap U$ with $y\ne x$.
Since an open set is a neighborhood of each of its points, no point of an open set can be a limit point of $A$. How can I prove this and what will happen if it were finitely many points?
 A: Suppose there is a neighborhood of $x$ that contains only finitely many points of $A$. If you consider the distances from $x$ to these finitely many points, can you find an open set containing $x$ and no other points of $A$?
For the other direction, I think writing down the negation of "$x$ is a cluster point of A" will be helpful.
A: Asaf Karagila and Brian M. Scott have already answered this, but perhaps the following will be a little clearer.
First, as Brian M. Scott has observed, the "if" half of the proof is immediate.
For the "only if" half of the proof, note that it suffices to prove every open ball $B(x,r)$ contains infinitely many points of $A.$
Thus, to prove the "only if" half, let $B(x,r)$ be given and assume that $x$ is a cluster point of $A.$ Using the definition of "$x$ is a cluster point of $A$" with $U = B(x,r),$ it follows that there exists $a_{1} \in B(x,r) \cap A$ such that $||x - a_1 || > 0.$
Let $r_1 = ||x - a_1||.$ Using the definition again, it follows that there exists $a_{2} \in B(x,r_1) \cap A$ such that $||x - a_2 || > 0.$
Let $r_2 = ||x - a_2||.$ Using the definition again, it follows that there exists $a_{3} \in B(x,r_2) \cap A$ such that $||x - a_3 || > 0.$
Continuing in this way shows that $B(x,r)$ contains infinitely many points of $A,$ namely the points $a_1,$ $a_2,$ $a_3,$ $\dots$ .
You'll want to explain how we know that $\{a_1,\;a_2,\;a_3,\;\dots\;\}$ is an infinite set. This follows from the fact that $k \neq k'$ implies $a_{k} \neq a_{k'},$ which in turn follows from the fact that $||x - a_1 || > ||x - a_2|| > ||x - a_3|| > \dots$ (i.e. $a_1,\;a_2,\;a_3,\;\dots$ are not only all different from each other, but in fact they all lie on different spheres centered at $x$).
A: Let’s get a misconception out of the way first. You wrote:

Since an open set is a neighborhood of each of its points, no point of an open set can be a limit point of $A$.

This is certainly not true: e.g., every point of the open set $(0,1)$ is a limit point of $(0,1)$ in $\Bbb R$. The other two limit points of $(0,1)$ are $0$ and $1$, and each open nbhd of each of them contains infinitely many other limit points of $(0,1)$.
You want to show two things:


*

*If $x$ is a cluster point of $S$, then every open nbhd of $x$ contains infinitely many points of $S$.    

*If every open nbhd of $x$ contains infinitely many points of $S$, then $x$ is a cluster point of $S$. 


The second one is trivial: an open nbhd of $x$ that contains at least two points of $S$, never mind infinitely many, is guaranteed to contain a point of $S$ different from $x$ itself. 
For the first one, let $U$ be an open nbhd of $x$. Choose a point $x_0\in(U\cap S)\setminus\{x\}$. Let $U_0=U\setminus\{x_0\}$ $U_0$ is an open nbhd of $x$, so you can choose a point $x_1\in(U_0\cap S)\setminus\{x\}$. Let $U_1=U_0\setminus\{x_1\}$. Do you see that this recursive construction can be extended to produce an infinite subset $\{x_n:n\in\Bbb N\}$ of $U\cap S$?
Added: The idea behind the argument that I suggested for (1) is important enough that you should learn it. In this case, to be sure, you can avoid it by proving the contrapositive instead: if $x$ has an open nbhd containing only finitely many points of $S$, then $x$ is not a cluster point of $S$. You need only find an open ball centred at $x$ that contains no point of $S\setminus\{x\}$. If $U$ is an open nbhd of $x$ such that $U\cap S=\{x_1,\dots,x_n\}$ is finite, can you find an $r>0$ such that $B(x,r)\cap(S\setminus\{x\})=\varnothing$? Warning: You cannot simply let $r=\min\{d(x,x_1),\dots,d(x,x_n)\}$. First, this may be $0$; why? Secondly, $B(x,r)$ may contain points of $S\setminus\{x\}$, even though it doesn’t contain $x_1,\dots,x_n$; why?
A: Recall that $\mathbb R^p$ is Hausdorff, which means that given $u,v$ we can find disjoint open sets $U,V$ such that $u\in U, v\in V$.
Let $x$ be a cluster point of $A$, take $U_0$ to be an open environment of $x$, then there is some $y_0\in A\setminus\{x\}$ which is in $U_0$.
Suppose that $U_n$ was chosen and $y_n\in U_n\cap A$ witnesses that $x$ is a cluster point of $A$. Let $U_{n+1}=B(x,r_{n+1})$ be an open ball around $x$ such that:


*

*$y_n\notin U_{n+1}$;

*$U_{n+1}\subseteq U_n$; and

*$r_{n+1}<2^{-n}$. 


We can make such choice because $U_n$ is open and therefore such an open ball exists. Now take $y_{n+1}\in U_{n+1}$ to be some element of $A$ which is not $x$.
Now I claim the following holds:


*

*$r_n\to 0$ and therefore $\bigcap U_n=\{x\}$;

*If $n>k$ then $y_n\in U_k$.

*If $n\neq k$ then $y_n\neq y_k$ (follows from 2).


These things are quite trivial to prove, I will skip those now.

Now let $U$ be an open set such that $x\in U$, therefore there is some $r>0$ for which $B(x,r)\subseteq U$. Let $n$ be such that $r_n<r$ then $B(x,r_n)=U_n\subseteq B(x,r)\subseteq U$.
From the second point above, the infinite set $\{y_k\mid k>n\}\subseteq A\cap B(x,r)\subseteq A\cap U$. As wanted.
