If (X,T) is perfect and A is a dense subset of X, then A has no isolated points. If $(X,T)$ is perfect and $A \subseteq X$ is a dense subset of X, then A has no isolated points.
Since $A$ is dense $\Rightarrow (\forall U \in T)(A \cap U \neq \emptyset)$ and
since $(X,T)$ is perfect $\Rightarrow$ $(\forall x \in X)(\{x\} \notin T)$ but  I can't figure a way to show that $(\forall x \in A) (\forall V \in T)(A \cap V \neq \{x\}) $.  It seems from that that all I can do is show that $A \cap U$ is nonempty.  I can't see how I'd show that it's $\neq \{x\}$.
 A: This is true if $X$ is $T_1$. Otherwise it need not be true: $X = \{0,1,2,3\}$, with topology $\left\{\emptyset, X, \{0,1\},\{2,3\}\right\}$ is perfect but $A = \{0,2\}$ is dense and consists of two isolated points. For a $T_0$ example, consider $X = [0,\infty)$ in the topology generated by all sets $[0,a) ,a > 0$. Here $X$ is perfect but $\{0\}$ is dense and trivially isolated. So I'll assume $X$ is $T_1$.
Suppose $x$ were an isolated point of $A$. So there is an open $U$ in $X$ such that $U \cap A = \{x\}$. But $\{x\}$ is not isolated in $X$, as $X$ is perfect. So pick $y \neq x$ with $y \in U$. As $X$ is $T_1$, we find an open set $V$ that contains $y$ but not $x$. But then $U \cap V$ is non-empty (it contains $y$), open, and it does not intersect $A$, as $(U \cap V) \cap A) = V \cap (U \cap A) = V \cap \{x\} = \emptyset$. This contradicts $A$ being dense. So $A$ has no isolated points.
A: $A$ is dense, suppose that $A$ has an isolated point $x$, this is equivalent to saying that $\{x\}$ is an open subset of $A$, $\{x\}=A\cap U$ where $U$ is an open subset of $X$. Remark that $U$ contains an element distinct of $x$ since $X$ is perfect. 
You also have $\{x\}$ is closed since it is isolated, thus $\{x\}=F\cap A$ where $F$ is a closed subset of $X$, $U-F$ is a non empty closed subset, but $(U-F)\cap A$ is empty contradiction. 
