Proof that a subset of the plane is an open set. I don't have a specific book that I am learning from because I am trying to learn on my own but my used definition of an open set $A$ is that for every $a\in A$ there exists an $\varepsilon>0$ such that the open ball $B(a,\varepsilon)\subset A$
Let $\boldsymbol{A}=\{(x,y)\in \mathbb{R}^2|0<x<1,0<y<2\}\subset \mathbb{R}^2$. Prove that $\boldsymbol{A}$ is an open set. 
My attempt: Let $P=(a,b)\in \boldsymbol{A}$. We can choose a radius $\varepsilon<\min\{|a-1|,|b-2|,a,b\}$. (This choice because they are the distances from the lines that bound the set $\boldsymbol{A}$ and it certainly seems to work). Now I think I would have to prove that if I take a point $(x,y)\in B(P,\varepsilon)$, then $(x,y)\in \boldsymbol{A}$, but I don't see how I can do this. I have tried to say that $\sqrt{(a-x)^2+(b-y)^2}<\varepsilon$ and the tried to somehow show in cases that always $0<x<1$ and $0<y<2$ but it doesn't seem to lead anywhere. I also don't even know how much work is needed to finish this proof because I haven't really seen any examples done. 
Can someone help me finish this argument or give a more elegant one alternatively(should still be from the definition)?

Let $(x,y) \in B(P,\epsilon),$ then $$(a-x)^2+(b-y)^2<\epsilon^2.$$ In particular $$(a-x)^2<\epsilon^2 \text{ and } (b-y)^2<\epsilon^2.$$ This gives $$|a-x|<\epsilon<a,|a-1|.$$So $0<x<1.$ Similary $0<y<2$ and thus $(x,y)\in B(P,\epsilon).$
