The set of accumulation points of A is closed I'm having a bit of a hard time proving or disproving the following claim in general topology:
Let X be a topological space, A $\subseteq$ X, and B the set of accumulation points of A. Is B necessarily closed?
 A: no. you need $X$ to be $T_1$ as mentioned in the comments, and if it is $T_1$ so you also presented by @mookid with a proof of that case. 
here is a counter example for the general case: let $X=\{a,b,c,d\}$ and $\tau =\{\{a,b,c\},\{b,c,d\},\{b,c\},X,\emptyset \}  $ note that $\{b,c\}$ is the set of accumulation of the set's $ \{a,b,c\},\{b,c,d\}$ and $\{a,b\}$ is not colsed.
A: $x $ is an accumulation point iff, for each 
open set $U$, $x\in U\implies  U\cap A - \{x\}\neq \emptyset$.
Consider $V$ the set of points, not accumulation points.
Let $x\in V$. There is an open set $U$ such as $$
x\in U\text{ and } U\cap A - \{x\} = \emptyset$$
In particular, $U\cap A\subseteq \{x\}$ so
if $y\in U$ then
$$
y\in U\text{ and } U\cap A - \{y\} = \emptyset
$$as well. Hence $V$ is open.
A: Here is another example using the overlapping interval topology http://en.wikipedia.org/wiki/Overlapping_interval_topology:
Let $X=[-1,1]$ with the topology whose open sets are $X$, $\emptyset$ and sets of the form $[-1,b), (a,b), (a,1]$ with $a<0<b$. In particular, note that $0$ is in every open set.
Take $A=\{0\}$. Let $x \in [-1,1]\setminus \{0\}$ and $x \in U$, with $U$ open. Since $0 \in U$ for all $U$ open, we see that $x$ is an accumulation point of $A$. Hence the accumulation points are $[-1,1]\setminus \{0\}$. The set $A$ is not open, hence the set of accumulation points is not closed.
