In a first-countable space, an accumulation point of the set of terms in a sequence is also a limit-point of the sequence I ran into something which seems like it should be true (and must be true because it is used in the proof of a theorem).
I've extracted the detail that I cannot quite verify.
Let $(x_{n})$ be a sequence in a first countable topological space $X$.  Let $y$ be an accumulation point of the set $I:=\{x_{n}:n\geq 1\}$ (that is, every neighborhood of $y$ contains $x_{n}$ for some $n\geq 1$ with the property that $y\neq x_{n}$).
Then $y$ is a limit point of the sequence $(x_{n})$.  That is, every neighborhood $U$ of $y$ has the property that $U\cap I$ is infinite.
Note: I'm not sure if First countability is required here, but it was a hypothesis in the theorem (and was indeed used elsewhere in the proof so it probably isn't required) so I included it just in case.
Why I am stuck: If $X$ is a metric space, the argument is simple, assume for a contradiction that the set $U\cap I$ is finite, then take a ball $B:=B(y,\epsilon)$ with $\epsilon$ equal to half of $min\{d(y,x_{n}) : x\in U\cap I, x\neq y\}$.  Then $U\cap B$ is a neighborhood of $y$ which contains no points of $I$ except possibly $y$ itself, which is a contradiction.
But since I'm not dealing with a metric space, this argument does not apply.
But I think that I can come up with a more general argument if $\{x\}$ is a closed set for every $x\in X$ (and more importantly for me, its complement is open).  
Is this normally a hidden assumption in topological setings?  Or is there another argument?
 A: Say $X = \{a,b,z\}$, where open sets are $\emptyset$, $X$, $\{a,b\}$, $\{z\}$.  So this is certainly not a $T_1$ space.  But it is first-countable, since there are only 4 open sets in all.  Note every neighborhood of $a$ contains $b$.  Here is our sequence: $x_1 = b$, and $x_k = z$ for all $k \ge 1$.  Then, according to your definition, $a$ is an accumulation point of the set $I$.  But $a$ is not a limit point of the sequence.
A: The general argument for a $T_1$-space $X$ goes like this. Suppose suppose that $x$ is an accumulation point of some set $S\subseteq X$. Let $U$ be any open nbhd of $x$, and suppose that $U\cap S$ is finite. Then for each $y\in U\cap S$ there is an open nbhd $B(y)$ of $x$ that does not contain $y$. Now let $$V=U\cap\bigcap_{y\in U\cap S}B(y)\;;$$ $V$ is the intersection of finitely many open nbhds of $x$, so $V$ is an open nbhd of $x$, and clearly $V\cap S\subseteq\{x\}$. This contradiction shows that $U\cap S$ must be infinite for each open nbhd $U$ of $x$.
Note that first countability is not in fact necessary, and neither is countability of $S$. However, as you thought, you do need to be able to separate $x$ from individual points of $S$ by open sets. Without that, there might be a point $y\in S\setminus\{x\}$ such that every open nbhd of $x$ contains $y$; if in addition $x$ has a nbhd $U$ that excludes all other points of $S$, which is certainly possible, $x$ will be an accumulation point of $S$, but $U\cap S$ will be finite. A simple example is to let $X=\{0,1\}$ with the topology whose open sets are $\varnothing,\{0\}$, and $X$: in this space $1$ is an accumulation point of $\{0\}$!
A: I found the (other) answer in another book:  Introduction to Topological Manifolds, by John M. Lee.
Using a nested sequence open sets (which is guaranteed by first countability, so it is actually needed!) about the point $y$, we immediately get a sequence which converges to $y$.
Of course the actual question I asked was about the closed-ness of singleton sets, which was answered already by Brian Scott and GEdgar (thanks!).
A: In Elementary Topology Problem Textbook, by O. Ya. Viro, O. A. Ivanov,
N. Yu. Netsvetaev, V. M. Kharlamov, 17.A.1, "Prove that in a space satisfying the first separation axiom a point is an accumulation point iff it is a limit point." Since 17.A.1 follows from 17.A, "If a first countable space is compact, then it is sequentially compact.", I thought writers wrongly replaced "first countable axiom" with "first separation axiom", but I was wrong, and I have proved that 17.A.1 is right. And considering the counter-example in the above answers, 17.A.1 might be the initial theorem you want to prove.
