Prove that If $x$ is an accumulation point of $S$, then every neighborhood of $x$ contains infinitely many points of $S$ Let $S \subseteq \mathbb{R}$.
To prove that every neighborhood of $x$ contains infinitely many points of $S$ whenever $x$ is an accumulation point of $S$, we will suppose to the contrary that there exists a neighborhood $N$ of $x$ such that 
$$N \cap S = \{s_1, \cdots, s_n\}$$
Since $x$ is an accumulation point of $S$, then every deleted neighborhood $N^*_{\delta} \cap S \neq \varnothing$. We then select a particular $\delta = \min\{|x-s_i| \, : \, s_i \neq x\}$. Then $\delta > 0$, but we have
$$N^*_{\delta} \cap \, (N \cap S) = \varnothing$$
which is a contradiction, so every neighborhood of $x$ contains infinitely many points of $S$.
Are there any problems with this proof?
 A: Let $B_\alpha$ be a neighborhood of $x$. That is:
$$ B_\alpha = \{z : |x - z| < \alpha\},$$
for $\alpha > 0$.
Take the sequence $$a_n = x + \frac{1}{n}.$$ This converges to $x$ as $n$ approaches to $+\infty$. This means that $x$ is accumulation point for some sets that contain $s$. Moreover:
$$\forall \varepsilon > 0 \exists m \in \mathbb{N} : n > m \Rightarrow a_n \in B_\varepsilon. $$
Now, consider the sets $S = B_\varepsilon$, $N = B_\alpha$ and $S \cap N$. It's clear that you can always find a $m'$ such that 
$$n > m' \Rightarrow a_n \in S \cap N,$$
and the $a_n$'s with $n > m'$ are, of course, infinite.
A: An alternative proof is the following that doesn't use contradiction.
Let $x$ be an accumulation point of $S$ 
Let $N$ be a neighbourhood of $x$.
We will construct a sequence $(x_n)$ of distinct points in $N\cap S$.
Let $\epsilon_1 >0 $ such that $B_{\epsilon_1} (x) \subset N$
This can be done since $N$ is a neighbourhood of $x$
Since $x$ is an accumulation point of $N$,
$\exists x_1 \in (B_{\epsilon_1}(x) \{x\}) \cap S$
Let $\epsilon_2 =\min \{ 1/2, d(x_1,x)\} >0$
As before, $\exists x_2 \in (B_{\epsilon_2}(x) \{x\}) \cap S$
Let $\epsilon_3 =\min \{ 1/3, d(x_2,x)\} >0$
Again, $\exists x_3 \in (B_{\epsilon_3}(x) \{x\}) \cap S$
And so on.
This way we have constructed a sequence of real numbers $(\epsilon_n)$ and $(x_n)$ in $S$ such that $0< \epsilon_{n+1} < \epsilon_n$
Thus $\forall n,m : n\neq m \Rightarrow x_n \neq x_m$ and $\forall n\geq 1, x_n \in B_{\epsilon_n} (x) \subset B_{\epsilon_1} \subset N$
This shows that $N\cap S$ contains infinitely many distinct points in $S$
