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> Show that $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\}$ is open in $\mathbb{R}^2$.
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The set can be expressed as $$A = \{(x,y) \in \mathbb{R}^2 : x^2+y^2< 2y\} \\
=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\} = \{(x,y) \in \mathbb{R}^2 : x^2+(y-1)^2 < 1\} = B((0,1),1)$$

and since open balls are open sets $A$ would be open. However I'm trying to use the metric to show this. If $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2 -2y< 0\}$ and I let $(x,y)\in A$ and $B((x,y), r)$, where $r=\min \{|x|,|y|\}$. Now picking $(x',y') \in B((x,y),r)$ I would want to show that $B((x,y), r) \subset A$, but I'm not entirely sure how to do this?