Prove that $S=\left \{ (x,y)\in \mathbb{R}^2:x>0\, \; or \; y>0\left. \right \} \right. \subset \mathbb{R}^2 $is open Prove that $S=\left \{ (x,y)\in \mathbb{R}^2:x>0\, \; or \; y>0\left.  \right \}  \right. \subset \mathbb{R}^2 is open$
I tried to choose r>0 such that $B(x,r)\subset S$
Let $y \in B(x,r)$ 
I want to prove $y \in S$ 
take $r<d(x,0)$
$d(x,y)<r<d(x,0)$
Geometrically i can see this proved it but i get stuck in the inequality 
so i need someone to complete my solution or get another one 
 A: Let $(x_0,y_0)$ be an arbitrary point in $S$. Let $r=\min\{|x_0|,|y_0|\}$. Then the open ball centered at $(x_0,y_0)$ of radius $r/2$ is in the interior of $S$.
A: Using the definition: Fix $(x,y)\in S$. Easiest way to start is to divide this situation to three separate cases: $(i)$ $x=0$; $(ii)$ $y=0$; and $(iii)$ $x\neq 0$ and $y\neq 0$. Note that for any $r>0$ we have $$B((x,y),r)\subseteq (x-r,x+r)\times (y-r,y+r),$$
i.e. we can fit the ball inside the above open rectangle.


*

*$(i)$ If $x=0$ then $y>0$ and for any $(z,w)\in B((x,y),y)$ we have $$(z,w)\in B((x,y),y)\subseteq (-y,y)\times(y-y,y+y)=(-y,y)\times (0,2y),$$ so $-y<z<y$ and $0<w<2y$. Since $w>0$ then $(z,w)\in S$. Thus $B((x,y),y)\subseteq S$. 

*The case $(ii)$ is analogous to the case $(i)$.

*$(iii)$ Assume then that $x\neq 0$ and $y\neq 0$. Then $r=\mathrm{min}(|x|,|y|)>0$, so for any $(z,w)\in B((x,y),r)$ we have $$(z,y)\in B((x,y),r)\subseteq (x-r,x+r)\times (y-r,y+r),$$
whence $x-r<z<x+r$ and $y-r<w<y+r$. We know that $x>0$ or $y>0$: assume without loss of generality that $x>0$ (the other case is analogous). Now $x\geq r$, whence $$0=x-x\leq x-r<z.$$
Since $z>0$ then $(z,w)\in S$. If we would have assumed that $y>0$ then it would have resulted analogously to $w>0$. So all together, $B((x,y),r)\subseteq S$.
By $(i)$, $(ii)$, and $(iii)$ you see that for any $(x,y)\in S$ there exists $r>0$ so that $B((x,y),r)\subseteq S$. Hence $S$ is an open set.
Alternative way: Recall that if $f:X\to Y$ is a continuous function then $f^{-1}(A)$ is an open subset of $X$ for every open subset $A$ of $Y$.
Now rewrite your set $S$ as 
\begin{align*}
S &=\{(x,y)\in\mathbb{R}^{2}:x>0\,\,\mathrm{or}\,\,y>0\} \\
&=\{(x,y)\in\mathbb{R}^{2}:x>0\}\cup\{(x,y)\in\mathbb{R}^{2}:y>0\} \\
&=\mathrm{pr}_{1}^{-1}(0,\infty)\cup\mathrm{pr}_{2}^{-1}(0,\infty),
\end{align*}
where pr$_{i}:\mathbb{R}^{2}\to\mathbb{R}$ is the projection map to the $i$:th coordinate. And conclude.
