Closed subset and accumulation points In general Topology by Kelley (1955), Theorem 5 states:

A subset of a topological space is closed if and only if it contains the set of its accumulation points.

a point $x$ is an accumulation point (aka limit point) of a subset $A$ of a topological space $(X, \tau)$ if and only if every neighborhood of $x$ contains points of $A$ other than $x$.
I'm trying to understand the theorem 5.
Suppose $a,b,c \in A$ and $A$ is closed in topological space $(X, \tau)$. Does it then imply that $\{b,c\},\{a,c\},\{a,b\}\in A$?
Edit
Okay, I got it. I did not understand definition of accumulation point correctly - I was under impression that it contains all points other than $x$ and does not contain $x$.
 A: It means that "you can't get infinitely close" to a point that's not in the set.
Example: the set $(0, 1]$ is not closed, because you can get as close to $0$ as you want while staying in the set (ie: in every open neighbourhood of $0$, you can find a point of $(0,1]$), but you couldn't reach it without leaving the set! However, if you now include $0$, the set $[0,1]$ is closed.
Other example: the unit disc in dimension $2$. If you include the boundary, then your set is closed, because you "contain" everything. But if you remove the boundary (ie, if you only consider the open disc $D = \{(x,y) : x^2 + y^2 < 1\}$), then you'll be able to get points to converge to the boundary, but could not reach it. 
This is what we commonly call closure!
In a general topological space, a set $A$ is closed iff it contains all its points of closure: ie the set of points $x$ such that for every neighbourhood $\mathcal{U}$ of $x$: $\mathcal{U}\cap A \ne \emptyset$.
note that this notion is slightly different from limit point, since we do not exclude the case where $\mathcal{U}\cap A$ is simply $\{a\}$. The equivalence "$A$ is closed $\iff A$ contains all its limits points is only true in certain types of topological spaces: they need to be first-countable (but all metric spaces do satisfy this property, so altogether it's not a bad way to represent it to yourself).
And $a, b, c \in A$ in your example are a finite example, so it doesn't really tell us much. However, say you take a converging sequence $(a_n$) in $A$, then this property tells you that its limit will also be in $A$
A: No. 
The sets you mention are in that case subsets of $A$ (not necessarily elements of $A$). 
This fact however is direct and has no connection at all with "$A$ being closed". 

To take away some possible misunderstandings:
A subset of a topological space is closed if and only it it contains the set its accumulation points as a subset. 
Or equivalently:
A subset of a topological space is closed if and only each of its accumulation points is an element of the set. 
A: Let $A$ be closed. Suppose $x$ is a limit point of $A$, we need to show that $x \in A$. Suppose $x \notin A$. Then $x \in O := X\setminus A$, which is open as $A$ is closed. But $O \cap A = \emptyset$ by definition, so this contradicts that $x$ is a limit point of $A$. This shows one implication.
Suppose that $A$ contains all its limit points, or $A' \subseteq A$ (where by definition $A'$ is the set of limit points of $A$ (could be empty, even). Now suppose $A$ were not closed, so $X\setminus A$ is not open, which means that there is an $x \in X\setminus A$ such that $x$ is not an interior point of $X\setminus A$, so that for all open neighbourhoods $O$ of $x$ we have $O \nsubseteq X\setminus A$, so for all open neighbourhoods of $x$ we have $O \cap A \neq \emptyset$. These intersection points are then clearly in $A\setminus \{x\}$, not just in $A$ (as $x \notin A$) and so $x \in A'$ by definition, but this contradicts the starting assumption $A' \subseteq A$, so our assumption that $A$ was not closed is false and $A$ is closed.
