Prove that $A=\{(x,y): x\leq 0, y\leq 0\}$ is a closed set in $\mathbb{R}^2$ 
Prove that $A=\{(x,y): x\leq 0, y\leq 0\}$ is a closed set in $\mathbb{R}^2$

So I want to prove that $A^c=\{(x,y): x>0 \text{  or  } y>0\}$ is open.
So if $(a,b)\in A^c$ then either $a>0$ or $b>0$
If $a>0$ then let $\epsilon = \frac{a}{2}$
then I want to show that $B_{\frac{a}{2}}((a,b))\subseteq A^c$
Let $(p,q)\in B_{\frac{a}{2}}((a,b))$
then $\vert\vert (p,q)-(a,b)\vert\vert<\frac{a}{2}$
then $\vert\vert (a,b)\vert\vert-\vert\vert(p,q)\vert\vert<\frac{a}{2}$
then $\vert\vert (a,b)\vert\vert -\frac{a}{2} <\vert\vert (p,q)\vert\vert$
I want to show that $p>0$, which intuitively I understand because geometrically the closest $(p,q)$ can be to the $y-axis$ must still have $p>\frac{a}{2}>0$
But I'm not sure how to show this.

$\|(a,b)-(p,q)\|=\sqrt{(a-p)^2+(b-q)^2}< \epsilon$ implies that $|a-p|< \epsilon={a\over 2}$, we deduce that $a-p< {a\over 2}$ and $a-{a\over 2}<p$ and ${a\over 2}<p$.
