Let $X$ be a metric space $\mathcal{G}$ an open subset of $X$ and $F \subset \mathcal{G}$ finite. Show that $\mathcal{G} \setminus F$ is open.
Let $x \in \mathcal{G} \setminus F$. This implies that $x \in \mathcal{G}$ but $\mathcal{G}$ was open by the problem statement and thus $\exists r >0$ such that $B(x,r) \subset \mathcal{G}$. So for arbitary $x \in \mathcal{G} \setminus F$ we can find an $r$ for which $B(x,r) \subset \mathcal{G} \setminus F$ so $\mathcal{G} \setminus F$ is open? I'm not sure I got this right is it true that if $x \in \mathcal{G} \setminus F$, then only possible scenario is that $x \in \mathcal{G}$?