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 $\neq \{x\}$.

$A$ 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. 
