Let $A$ be a set, $X$ a metric space, $x$ an accumulation point of $A$ that is, every nbhood of $x$ contains a point $a \in A$, $a \neq x$.
I wrote a proof of the fact "every nbhood of $x$ contains infinitely many points of $A$". But for one I don't know if it is correct and also if it is correct whether it can be improved. Here is my proof:
The open ball $B(x,\varepsilon)$ is a neighbd of $x$ and by assumption contains $a \in A, a \neq x$. Since $B(x, \varepsilon)$ is open there is $\delta$ such that $B(x, \delta) \subseteq B(x, \varepsilon)$ and $a \notin B(x, \delta)$. Since $B(x, \delta)$ is a nbhood of $x_0$ by assumption there is $a' \in A$ with $a' \in B(x, \delta)$.
Then I wrote: "Repeat process to obtain infinitely many points $a \in A$ in $B(x, \varepsilon)$".
Is this valid? And if it is not: how to finish the proof? And if it is correct: is there a neater way to write this proof? Thank you.