$\{ \left< x,y \right> \in \mathbb{R}^2 \mid ( \frac{x}{3} )^2 + (\frac{y}{4})^2 \lt 1 \}$ is open 
$S=\left\{ \langle x,y \rangle \in  \mathbb{R}^2 ~\middle|~ \left( \frac{x}{3} \right)^2 + \left(\frac{y}{4}\right)^2 \lt 1 \right\}$ is open in $\mathbb{R}^2$

I have just started reading about open/closed sets and this is one of the first problems I came across. I think I understand the definition properly, but I need help in getting started with proving these kinds of problems. For e.g., in $\epsilon-\delta$ proofs, the trick is to find $\delta$ in terms of $\epsilon$ by working your way backwards from a desired result.
I am trying to do the same here but I seem to get no where.
For these kinds of problems, here is what I need to do:
$\forall \textbf{x} \in S$, I must find $r \gt 0$ such that $B_r(\textbf{x}) \subseteq S$.
More specifically,
Given $\textbf{x} \in S$, find $r \gt 0$ such that if we let $\textbf{y} \in B_r(\textbf{x})$, then $\textbf{y} \in S$.
That is 
$\textbf{x}=\langle x_1,x_2\rangle, \textbf{y}=\langle y_1,y_2 \rangle$
we are given $\left( \frac{x_1}{3} \right)^2 + \left(\frac{x_2}{4}\right)^2 \lt 1$ 
and
$\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \lt r$
we need to prove
$\left( \frac{y_1}{3} \right)^2 + \left(\frac{y_2}{4}\right)^2 \lt 1$ 
I know, I need to come up with some clever $r$ in terms of $\textbf{x}$. 
I have been trying to work backwards so as to naturally "see" an $r$ that would work. But I am lost.
I would like a nudge in the right direction.
 A: As you've noticed it's complicated to work with ellipses. But notice that
$$\left(\frac{x}{4}\right)^2 + \left(\frac{y}{3}\right)^2 \geq \left(\frac{x}{4}\right)^2 + \left(\frac{y}{4}\right)^2$$
Use this so you can work with circles instead to find an $r$.
Edit: Another hint. Use AM-GM inequality to get 
$$\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \geq \sqrt{2|x_1-y_1||x_2-y_2|}$$
Given that the points live in an ellipse, can you put a bound on the largest possible horizontal and vertical distances between points?
A: If you want to find explicitly for each $(x,y) \in S$ a ball $B_{r(x,y)} \subset S$ you may proceed as follows:


*

*For a given $(x,y) \in S$ set $R = \left( \frac{x}{3} \right)^2 + \left(\frac{y}{4}\right)^2 < 1$

*Now, you may parametrize a ball with radius $r$ around $(x,y)$ using polar coordinates $(x+r\cos t, y + r \sin t)$

*Plug this into the equation of the ellipse
$$\left(\frac{x+r\cos t}{3} \right)^2 + \left(\frac{y+r\sin t}{4}\right)^2$$
$$ =\left( \frac{x}{3} \right)^2 + \left(\frac{y}{4}\right)^2 + \frac{2xr\cos t + r^2\cos^2t}{9} + \frac{2yr\sin t + r^2\sin^2 t}{16}$$
$$= R + r\underbrace{\left(\frac{2x\cos t + r\cos^2t}{9} + \frac{2y\sin t + r\sin^2 t}{16} \right)}_{g(r,t):=}\stackrel{!}{<}1$$

*Now, estimate $|g(r,t)|$ roughly by a constant. For example
$$|g(r,t)| \stackrel{r<1,|x|<3,|y|<4}{<} \frac{7}{9}+ \frac{9}{16}<2$$

*Finally, choose $0<r<1$ such that $r <\frac{1}{2}\min(R,1-R)$. So, you have $r|g(r,t)|<R$ and $r|g(r,t)|<1-R$. Hence
$$0 < R+rg(r,t) < 1 \Rightarrow B_r(x,y)\subset S $$ $$\mbox{ for } 0 <r <\frac{1}{2}\min\left(\left( \frac{x}{3} \right)^2 + \left(\frac{y}{4}\right)^2, 1- \left( \left( \frac{x}{3} \right)^2 + \left(\frac{y}{4}\right)^2\right)\right)$$
