Show subset of $\mathbb{R}^2$ is open Let $D = \{ (x,y) \in \mathbb{R}^2: -1 < x < 1 \text{ and } 0 < y < \sqrt{1-x^2} \}$.
Show that $D$ is an open subset of $\mathbb{R}^2$ using the definition of an open set.
Definition: The subset $D \subseteq \mathbb{R}^2$ is said to be open if every point in $D$ is an interior point of $D$. 
Or equivalently, $D =$ int($D$) $= \{a \in D: \exists r > 0 : B(a;r) \subseteq D \}$, where $B(a;r) := \{ b \in \mathbb{R}^2 : ||b-a|| < r\}$.
So I should show that $D =$ int($D$). But I find the definition of $D$ hard to work with, not sure what a proper $r$ would be. 
 A: For a point $x,y$ let $s$ be the minimum of the distance from $x,y$ to the circle $x^2+y^2=1$ and the distance to the line $y=0$. By assumption, $s>0$. Then note that $B((x,y), s/2)\subseteq D$ is open.
A: First step is to understand how you set $D$ looks like.

 $$D=\{(x,y)\in\mathbb{R}^2~:~x\in[-1,1],~0<y<\sqrt{1-x^2}\}=\{(x,y)\in\mathbb{R}^2~:~x^2+y^2<1,~y>0\}$$ is the upper half of the open disk with center $(0,0)$.

Now you can see, how to choose the radius such that you remain in $D$.
Solution:
Choice of $r$:

 For $(x,y)\in D$ you have to choose $r<1-\|(x,y)\|$ to remain in the open disk and $r<y$ to remain in the upper half of the disk. Therefore define $r<\min\{1-\|(x,y)\|,y\}$. 

Proof of $B((x,y),r)\subset D$:

 If $(v,w)\in B((x,y),r)$ then you have \begin{align}v^2+w^2&=\|(v,w)\|=\|(v,w)-(x,y)+(x,y)\|\leq\|(v,w)-(x,y)\|+\|(x,y)\|\\&<r+\|(x,y)\|<1-\|(x,y)\|+\|(x,y)\|=1\end{align} and $$|w-y|= \sqrt{(w-y)^2}\leq \|(v,w)-(x,y)\|<r<y$$ and therefore $-y<w-y<y$ which yields $0<w$. All in all we got $(v,w)\in D$ and $B((x,y),r)\subset D$.

