Is the following argument correct?
Let $D_1$ and $D_2$ be any open discs in $\mathbf{R}^2$ with $D_1\cap D_2\neq\varnothing$. If $(a,b)$ is any point in $D_1\cap D_2$, show that ther exists an open disc $D_{(a,b)}$ with center $(a,b)$ such that $D_{(a,b)}\subset D_1\cap D_2$.
Proof. Let $D_1,D_2$ denote two arbitrary discs in $\mathbf{R}^2$ having $\alpha = (a_1,b_1)$ and $\beta = (a_2,b_2)$ as there centers
and radii $r_1$ and $r_2$ respectively, that is \begin{align*} D_1 =
\{(x,y)\in\mathbf{R}^2:(x-a_1)^2+(y-b_1)^2<r_1^2\}\\ D_2 =
\{(x,y)\in\mathbf{R}^2:(x-a_2)^2+(y-b_2)^2<r_2^2\} \end{align*} Let
$\gamma = (a,b)\in D_1\cap D_2$. We define the disc $D_{(a,b)}$
similar to $D_1,D_2$ but having center $(a,b)$ and radius $r =
\min\{\frac{r_1-d_1}{8},\frac{r_2-d_2}{8}\}$, where $d_1$ and $d_2$
are defined as follows
\begin{align*} d(\gamma,(\alpha,\gamma)) = d_1
= \sqrt{(a-a_1)^2+(b-b_1)^2}\\ d(\gamma,(\beta,\gamma)) = d_2 = \sqrt{(a-a_2)^2+(b-b_2)^2}
\end{align*} Now let $(x,y)$ be an
arbitrary point inside the disc $D_{(a,b)}$, appealing to the triangle
inequality then yields
\begin{align*} \sqrt{(x-a_1)^2+(y-b_1)^2}
&=d(\alpha,(x,y))\leq d(\alpha,\gamma)+d(\gamma,(x,y))\\
&=d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1
\end{align*}
\begin{align*} \sqrt{(x-a_2)^2+(y-b_2)^2} &= d(\beta,(x,y))\leq
d(\beta,\gamma)+d(\gamma,(x,y))\\
&=d_2+r<d_2+\frac{r_2-d_2}{8}=\frac{r_2}{8}<r_2 \end{align*} Since our
choice of $(x,y)$ was arbitrary, we have $D_{(a,b)}\subset D_1\cap D_2$.$\blacksquare$
Best Answer
The proof is correct and mostly good, there are only minor mistakes and typos.
You write $$ d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{8}<r_1 $$ and here the equality is false. But this can be easily fixed, for example (using $d_1< r_1$) $$ d_1+r<d_1+\frac{r_1-d_1}{8}=\frac{r_1}{4}<r_1 . $$ The same issue exist with $d_2,r_2$ below.
In my opinion it would be nicer if you choose $$ r= \min\{r_1-d_1,r_2-d_2\}, $$ then your proof would still work, but your choice is not wrong!
Small typos: You probably mean to write $d(\gamma,\alpha)$ or $d(\gamma,(a_1,b_1))$ instead of $d(\gamma,(\alpha,\gamma))$ and $d(\gamma,\beta)$ or $d(\gamma,(a_2,b_2))$ instead of $d(\gamma,(\beta,\gamma))$.