# Intersection of two open sets is open

Proposition: Intersection of two open sets is open.

Proof: Let $$(X,d)$$ be a metric space and $$A_1, A_2$$ are non-empty open sets such that $$A_1, A_2 \subseteq X$$. Further let a point be $$x \in A_1 \bigcap A_2$$. Then, $$x \in A_1$$ and $$x \in A_2$$. As $$A_1$$ is open, $$\forall x \in A_1$$ there exists $$r_1>0$$ such that $$d(x,y) and $$y \in A_1$$. Conversely, $$\forall x \in A_2$$, there exists $$r_2>0$$ such that $$d(x,y) and $$y \in A_2$$. Thus, $$y$$ is an interior point of both $$A_1,A_2$$, implying that it also has to be the interior point of $$A_1 \bigcap A_2 \square$$

I wanted to check if my proof proves the proposition.

Looks like your ideas are right but the way you lay it out is confusing. $$x\in A_1$$ implies there exists $$r_1>0$$ such that whenever $$d(x,y), then $$y\in A$$. Then you can take $$r=\min\{r_1,r_2\}$$