Determine if the set $A=\{(x,y) \in \overline{B}\mid x \geqslant0 \}$ is open or closed. 
Let $\overline{B}= \{(x,y) \in \Bbb{R}^2 \mid x^2+y^2 \leqslant 1 \}.$ Determine if the set $A=\{(x,y) \in \overline{B}\mid x \geqslant0 \}$ is open or closed.

If I let $(x,y)\in A$ and $r >0$ then in order for $A$ to be open it would have to satisfy that for every $(x,y) \in A$ the open ball $B((x,y),r)$ would be a subset of $A$. I think this set is not open, but I'm not sure how to prove it? For example I have that $(-\frac{\varepsilon}{2},0) \in \overline{B}\setminus A$, but $(0,0) \in A.$

Yes, this shows that $A$ is not open in $\bar B$. To show that it is closed, note that $\bar B\setminus A=\{(x,y)\in\Bbb R^2:x<0\}\cap\bar B$ which is open in $\bar B$ since the former set is open in $\Bbb R^2$.
