**Proof Plan.** Given a nonempty open set $W$ and a [sequence](/page/Sequence) of dense [open sets](/page/Open%20Set) $\{G_n\}$, we construct a nested sequence of closed balls $\overline{B}(x_n, r_n)$ with $r_n \to 0$, each contained in $W \cap G_1 \cap \cdots \cap G_n$. The Cantor Intersection Theorem produces a point in all of them, hence in $W \cap \bigcap_n G_n$. **Step 1 (First ball).** Since $G_1$ is dense and $W$ is open nonempty, $G_1 \cap W \neq \varnothing$ and is open. Choose $x_1 \in G_1 \cap W$ and $r_1 > 0$ with $\overline{B}(x_1, r_1) \subseteq G_1 \cap W$ and $r_1 < 1$. **Step 2 (Induction).** Given $\overline{B}(x_n, r_n) \subseteq G_1 \cap \cdots \cap G_n \cap W$ with $r_n < 1/n$, the ball $B(x_n, r_n)$ is open and nonempty, and $G_{n+1}$ is dense, so $G_{n+1} \cap B(x_n, r_n) \neq \varnothing$ and is open. Choose $x_{n+1}$ and $0 < r_{n+1} < 1/(n+1)$ with $\overline{B}(x_{n+1}, r_{n+1}) \subseteq G_{n+1} \cap B(x_n, r_n)$. **Step 3 (Convergence).** The closed balls $\overline{B}(x_n, r_n)$ are nested and nonempty with diameters $\le 2/n \to 0$. By the [Cantor Intersection Theorem](/theorems/624), $\bigcap_{n=1}^\infty \overline{B}(x_n, r_n) = \{x\}$ for some $x \in X$. **Step 4 (Conclusion).** Since $\overline{B}(x_n, r_n) \subseteq G_n$ for each $n$, we have $x \in G_n$ for all $n$. Since $\overline{B}(x_1, r_1) \subseteq W$, we have $x \in W$. Therefore $x \in W \cap \bigcap_{n=1}^\infty G_n$, so this intersection is nonempty. Since $W$ was an arbitrary nonempty open set, $\bigcap_n G_n$ is dense. $\blacksquare$