Bolzano-Weierstrass: "Neighbourhood of x contains an infinite number of members from a sequence"? Accumulation point? I've not heard of these terms before, but was viewing a Bolzano-Weierstrass theorem proof that proved that the "Neighbourhood of x contains an infinite number of members from the sequence" in order to reach for the conclusion of a limit existing (in that neighbourhood).
The proof is here:
https://ram.rachum.com/bw.htm
What does it mean for a neighbourhood to contain an infinite number of numbers? What theorems is this related to (i.e. where does this "infinite number of numbers" notion come from?)?
 A: Neighbourhood of $x$ contains an infinite number of members from the sequence $a_n$ means that given $\epsilon>0$, there are infinite number of $a_{n_k}$ such that $a_{n_k}\in(x-\epsilon,x+\epsilon)$. It also means that $x$ is an accumulation point of $a_n$.
$$
\lim_{k\to\infty}a_{n_k}=x
$$
Next we give a topological proof that any neighbourhood of $x$ contains an infinite number of $A$ is equivalent to that $x$ is an accumulation point of $A$. 
First, we prove that if $x$ is an accumulation point of $A$, then any neighbourhood of $x$ contains an infinite number of $A$. 
Suppose not. Then there is a neighborhood of $x$, says $M$, which has only finite many points $a_1,…,a_n$ of $A$. 
By separation axiom of Hausdorff space, for each $a_k\in M$ and $x$ ($1\leqslant k\leqslant n$), there are open set $U_k$ and $V_k$ of $M$, $a_k\in U_k, x\in V_k$, such that $U_k\cap V_k=\varnothing$. Let $V=\bigcap \limits_{k=1}^{n}V_k$. Clearly $V$ is open set, and for each $k, 1\leqslant k\leqslant n$, $U_k\cap V\subset U_k\cap V_k$. And so $U_k\cap V=\varnothing$ or $a_k\notin V$. 
Therefore $V$ is an open subset of $M$ that contains $x$ but contains no point of $A$, which means $x$ is not a limit point of $A$, contradiction.
Second, we prove that if any neighbourhood of $x$ contains an infinite number of $A$, then $x$ is an accumulation point of $A$. 
Suppose not. Then $x$ is not a limit point of $A$. So there is an open set $O$ containing $x$ and contains no point of $A$ but $x$. But $O$ is a neighborhood of $x$. Thus $O$ has infinitely many points of $A$. This is contradiction.
