Show that $[-1,1] \times [-1,1]$ is a closed set. Show that $A = [-1,1] \times [-1,1]$ is a closed set.
I know that I have to show $A^c$ is open. So I have to find $\epsilon > 0$ 
sufficiently small such that for all $x \in A^c$, $B(x,\epsilon) \subset A^c$. I am a bit blocked at this point­. I think I have to use the triangle inequality and Cauchy-Schwarz.
Is anyone can give me a hint?

It is an easy exercise to show that the following are open subsets of $\mathbb{R}^2$ for any $r\in\mathbb{R}$:
\begin{align*}
(r,\infty)\times\mathbb{R} &&
(-\infty,r)\times\mathbb{R} &&
\mathbb{R}\times(r,\infty) &&
\mathbb{R}\times(-\infty,r) &&
\end{align*}
Now notice that $([-1,1]\times[-1,1])^C$ can be described as a union of sets of this type for certain appropriate choices of $r$.  Thus the complement is open as desired.
