My best guess for what $[b_1,b_2]$ is supposed to mean is the subring of $A$ generated by $b_1$ and $b_2$. However, even with this interpretation, there are some statements that aren't quite right (though as I recall, this particular paper has a lot of minor errors of this sort, so this shouldn't be too surprising). Here is how I would rewrite the final three sentences of the proof of Theorem 4:
Let $f:\mathbb{Z}[x_1,x_2]\to A$ be the unique homomorphism sending $x_i$ to $b_i$ and let $I=(x_1,x_2)\subset\mathbb{Z}[x_1,x_2]$. For any $z\in I$, $f(z)$ vanishes off of $d(b_1)\cup d(b_2)$, and its image restricted to $d(b_1)\cup d(b_2)$ is equal to its image restricted to the finite set $Y$. Thus we need only find $z\in I$ such that $z\not\in f^{-1}(\phi(y))$ for each $y\in Y$, and then $c=f(z)$ will be an element of $(b_1,b_2)$ such that $d(c)=d(b_1)\cup d(b_2)$. But each $f^{-1}(\phi(y))$ is a prime ideal in $\mathbb{Z}[x_1,x_2]$ which does not contain $I$ (since every point of $Y$ is in $d(b_1)\cup d(b_2)$), and there are only finitely many of them, so by prime avoidance such a $z$ must exist.