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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 $d(x,y)<r_1$ and $y \in A_1$. Conversely, $\forall x \in A_2$, there exists $r_2>0$ such that $d(x,y)<r_2$ 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.

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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 $d(x,y)<r_1$, then $y\in A$. Then you can take $r=\min\{r_1,r_2\}$

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