Union of a countable collection of open balls

($\rightarrow$) Let $Y \subset \mathbb{R}^n$ be open, then for every $x \in Y$, there exists an open $U_x \subset Y$ such that $x \in U$. Clearly the set $\left\{ \displaystyle\bigcup U_x \colon x \in Y \right\} = Y$. However the collection $\mathscr{F} := \left\{ U_x \colon x \in Y \right\}$ may not be countable. Since $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$, it is also dense in $Y$. Restrict $\mathscr{F}$ by considering then $\mathscr{G} := \left\{ U_x \in \mathscr{F} \colon x \in Y \cap \mathbb{Q}^n \right\}$, which is countable and covers $Y$ since $\mathbb{Q}^n \cap Y$ is dense in $Y$.

($\leftarrow$) Suppose that $Y \subset \mathbb{R}^n$ is the union of a countable collection $\mathscr{C}$ of open sets. By hypothesis, for every $x \in Y$, there exists some open set $U_x \in \mathscr{C}$ such that $x \in U_x$ and $U_x \subset Y$. Therefore by definition, $Y$ is open.

see: Math 104. Metric Spaces, Proposition 29.