$A^\prime$(the set of limit points of $A$) is connected. Suppose that every singleton in X is a closed set. A is a connected subset of X. Then  $A^\prime$(the set of limit points of $A$) is connected.
I can't use the conditions in the question well. I know we should use contradiction but I fail to prove it.I know in the $T_{1}$ space, $A^\prime$ is clsoed.And in if $A$ is connected then $A$ added some limit points is also connected. But they didn't work.So how to prove this, can someone give me some hints?
 A: First of all the closure of $A$ is always connected. This happened because if $\phi: cl(A)\to \{0,1\}$ is a continuous map, then it is continuous on $A$, that is connected, and so it is constant on $A$ (for example, its value is $0$). By contradiction, if there exists $x\in cl(A)\setminus A$ such that $\phi(x)=1$, then $x\in \phi^{-1}(1)$, i.e. $\phi^{-1}(1)$ is an open neighborhood of $x$. This means there exists $y\in A\cap \phi^{-1}(1)$ and so $\phi(y)=1$ that is a contradiction.
Now we can prove $A'$ is connected observing that $cl(A)=A'\cup I(A)$, where $I(A)$ is the set of isolated points of $A$. In our case $A$ has not isolated points because $A$ is connected. This means
$A'=cl(A)$ is connected.
Remark: You’re need the $T_1$ condition when you say $A$ has not isolated points. In fact if there exists an isolated point $x$, then there is an open neighbourhood $U$ of $x$ in $X$ such that $U\cap A=\{x\}$, i.e $\{x\}$ is an open and closed subset of $A$. This is a contradiction because $A$ is connected.
