Proof: An accumulation point if each neighborhood contains point not itself I want to confirm my proof of this lemma.
Lemma: Let $S$ be a set of real numbers. Then $a$ is an accumulation point of $S$ if and only if there each neighborhood of $a$ contains a member other than itself.
The forward direction is trivial and obvious.
Assume $a$ is an accumulation point then each neighborhood of $a$ has infinitely many points and so it contains a member other than itself.
It is the reverse direction I wish to confirm. 
Assume we have $a \in S$ where every neighborhood $Q$ of $a$ contains at least on point not $a$ but in $S$. Appealing to the density of the reals there are therefore infinitely many points in each neighborhood and so $a$ is an accumulation point.
Is everything correct here?
 A: If you want to do it rigourously, you can do it this way :
Let $a$ be a point such that every neighborhood contain at least one point different from $a$.
Take an arbitrary neighborhood $U$ of $a$. We will construct by induction a sequence of elements in $U$, all differents.
Step 1 :
$U$ is a neighborhood of $a$, so by hypothesis, it exists $x_1 \in U$, $x_1 \neq a$


*

*$x_1 \in U$

*$x_1\neq a \Rightarrow  |x_1-a|>0$ 


Step 2 :
Let's consider $U_2 = U \cap B(a,|x_1-a|)$ (where $B(x,r)$ is the open ball of center $x$ and radius $r$). It is a neighborhood of $a$, so it exists $x_2 \in U_2$, $x_2 \neq a$. 


*

*$x_2 \in U_2 \Rightarrow x_2 \in U$

*$x_2 \neq a \Rightarrow |x_2-a|>0$

*$x_2 \in B(a, |x_1-a|) \Rightarrow |x_2-a|< |x_1-a|$
So we have $x_1,x_2 \in U$ different from $a$ and between them that verify :
$0< |x_2 - a | < |x_1-a|$
Step n :
Suppose we have $n-1$ elements in $U$ that verify $0< |x_{n-1} - a | < \cdots < |x_1-a|$. Consider $U_n = U \cap B(a,|x_{n-1} - a |)$. It is a neighborhood of $a$, so it contain $x_n \neq a$.


*

*$x_n \in U_n \Rightarrow x_n \in U$

*$x_n \neq a \Rightarrow |x_n-a|>0 $

*$x_n \in B(a, |x_{n-1}-a|) \Rightarrow |x_n-a|< |x_{n-1}-a|$
We have $n$ elements in $U$ that verify $0< |x_n-a| < |x_{n-1} - a | < \cdots < |x_1-a|$
We can conclude by induction that there is an infinity of distincts elements in $U$
