In an exercise I have to prove the following:
Let $D_1,D_2 \in \mathbb R ^2$ be two open discs with $D_1 \cap D_2 \neq \emptyset$. If $(a,b)$ is any point in $D_1 \cap D_2$, show that there exists an open disc $D_{(a,b)}$ with center $(a,b)$ such that $D_{(a,b)} \subset D_1 \cap D_2$.
My approach:
Let $\bar D_1$ and $\bar D_2$ be the closure of the respective sets. Let $\partial(D_1\cap D_2)$ be the border of the intersection of $\bar D_1$ with $\bar D_2$. If we denote the $A=(a,b)$ then we can define: $\epsilon = \min \{\ \overline{XA},\ \forall X \in \partial(D_1\cap D_2)\}$.
If we define $D_{(a,b)}$ to be the open ball with center $(a,b)$ and with radius $\frac{\epsilon}{2}$, Then $D_{(a,b)} \subset D_1 \cap D_2$.
Is is this proof valid? If so,is this argument enough or do I need to prove or to add anything else? What other interesting ways there exists to prove this?
For the sake of curiosity, here it is an alternative solution based on the concept of distance.
Let $(X,d_{X})$ be a metric space. Then any open ball is open as it has been proved here. Based on it, we can solve the exercise proposed.
Let us consider two open balls $B(x_{0},\delta)$ and $B(y_{0},\varepsilon)$ whose intersection $B(x_{0},\delta)\cap B(y_{0},\varepsilon)$ is not empty.
Let $y\in B(x_{0},\delta)\cap B(y_{0},\varepsilon)$. Then we conclude there exist $B(y,r_{1})\subseteq B(x_{0},\delta)$ and $B(y,r_{2})\subseteq B(y_{0},\varepsilon)$ for some positive real numbers $r_{1}$ and $r_{2}$.
Without loss of generality, we can assume that $r_{1}\leq r_{2}$. Then $B(y,r_{1})\subseteq B(y,r_{2})\subseteq B(y_{0},\varepsilon)$.
Based on the previous arguments, we finally conclude that $B(y,r_{1})\subseteq B(x_{0},\delta)\cap B(y_{0},\varepsilon)$, and we are done.
Hopefully it is useful.