Validity of a proof requiring a non-terminating process? I was trying to prove directly that in a Hausdorff space, if $p \in A$ is a limit point of $A$, then every neighborhood of $p$ contains infinitely many points of $A$. The proof I remember seeing a while back drew a contradiction, but I wanted to try doing it more directly. 
I figured you could start with a neighborhood $U$ of $p$, and then use the limit point condition to get that it contains a point $a_1$ of $A$, then use the Hausdorff condition to get two disjoint neighborhoods $N_1,N_1'$ of $p$ and $a_1$, and then consider $N_1 \cap U$ which is also a neighborhood of $p$ disjoint from $N_1'$ (and still contained in $U$) and use the limit point condition again to get a new point $a_2$ inside $N_1 \cap U$ and then use the Hausdorff condition again, so on and so forth.
It seems reasonable enough, but it started sounding suspiciously like trying to use induction on an infinite set. Is that what this construction really is? I suppose that even though I feel that I am constructing something infinite, that at each stage I really only have finitely many objects, so I can never actually get to the part where I will have the infinite set that I am looking to make. So in general is it not valid to give the recipe for some non-terminating process and claim that it leads to a valid construction?
 A: Arguments that seem to involve an infinite process can often be rephrased as involving a notion of induction. In this case, I would prove the following statement by induction:

For every natural number $n$, every neighbourhood of a limit point $p$
  contains at least $n$ points from $A$.

The argument outlined in the question can be easily modified to show this.
(This will of course only establish that there are countably many such points from $A$ in every neighbourhood.)
A: It can be seen as a completely valid recursive construction of an infinite subset. Let's formalise it: 
Let $p$ be a limit point of the infinite set $A$ in a $T_1$ space $X$.
Let $O$ be an open neighbourhood of $p$. First pick $a_0 \in O \cap A$ with $p \neq a$, which is possible as $p \in A'$. 
We will define $a_n, n \in \mathbb{N}$ such that, at stage $n$:


*

*all $a_i, i \le n$ are pairwise distinct and all distinct from $p$.

*all $a_i \in A \cap O ,i \le n$.


Now we find $a_{n+1}$ as follows: note that $F = \{a_0,a_1,\dots,a_n\}$ is finite, so $X \setminus F$ is open in $X$ by $T_1$-ness, and is a neighborhood of $p$.
Then $O \cap (X\setminus F)$ is also a neighbourhood of $p$ so intersects $A \setminus \{p\}$, pick $a_{n+1} \in A \cap (X\setminus F) \cap (O \setminus \{p\})$. Then the first condition is met for $a_{n+1}$, as $a_{n+1} \notin F$ and by construction it's not $p$, and the second condition is also clear.
As stage $0$ was also trivially satisfied, standard recursion theorems (e.g. see this question) then tell us that $B = \{a_n: n \in \mathbb{N}\}$ is a well-defined set, and the first condition tells us that $B$ is infinite and the second that $B \subseteq A \cap O$, as required.
A: You have constructed an infinite sequence of distinct points,
inside U.  Thus U is infinite.
Exercise.  Show the proposition holds for all T1 spaces.
