If a space $X$ has no isolated points, then nor does any dense subset of $X$.

The claim is true assuming the $T_0$ (Kolmogorov) axiom.

Let $S\subseteq X$ be dense. And suppose $x_0\in S$ is isolated. Then there exists an open set $X_0\ni x_0$ such that $X_0 \cap S = \{x_0\}$. Let $x_1\in X_0$ and $N$ be an open neighbourhood of $x_1$. Suppose $x_0 \not\in N$, then $N \cap X_0 \subset X_0\setminus \{x_0\}$ is open, contradicting density of $S$. Therefore $x_0$ must be in every neighbourhood of $x_1$.

Therefore in $T_0$ holds $x_0$ is isolated in $X$

(The OP may want to check whether $T_0$ is a standing assumption in the book. It is even more reasonable to assume that than assuming Hausdorff.)

Editor's note: The above proof actually assumes $T_1$, see YCor's comment below for what went wrong and a counterexample. (I'm leaving this note as the answerer is currently unregistered)