I am attempting the proof but encounter a problem as described at the bottom.
Fix arbitrary $\epsilon>0$. Then need to show that there exists a $\delta>0$ such that if $0<\sqrt{(x-2)^2+(y-3)^2}<\delta$, then \begin{align*} |3x-2y+4|&=|3x-2y+4-6+6|\\ &=|3(x-2)+2(y-3)+4|\\ &\leq 3|x-2|+2|y-3|+4\\ &\leq 3\delta +2\delta +4=5\delta+4 \end{align*} We want $5\delta+4<\epsilon$ so let $\delta=\frac{\epsilon-4}{5}$. Above, we used that $|x-2|$ and $|y-3|\leq \sqrt{(x-2)^2+(y-3)^2}$
Now fix arbitrary $\epsilon>0$ and choose $\delta=\frac{\epsilon-4}{5}$. Then \begin{align*} 0<\sqrt{(x-2)^2+(y-3)^2}<\frac{\epsilon-4}{5}&\implies 5\sqrt{(x-2)^2+(y-3)^2}+4<\epsilon\\ &\implies 3|x-2|+2|3-y|+4<\epsilon\\ &\implies |3(x-2)+2(3-y)+4|<\epsilon\\ &\implies |3x-6+6-2y+4|<\epsilon\\ &\implies |3x-2y+4|<\epsilon \end{align*}
But how do we know that $\delta>0$? Say if $\epsilon<4$. How can I rectify this problem?