Proving that the unit square $\{(x,y) \in \mathbb{R^{2}} | 0Proving that the unit square $\{(x,y) \in \mathbb{R^{2}} | 0<x,y<1\}$ is open in the metric topology on $\mathbb{R^{2}}$
What I know so far is that for a Metric space with (X,d) where X is a set and d is the metric, the following properties must hold: 
i) $d(x,y) = 0$ 
ii) $d(x,y) = d(y,x)$
iii) $d(x,z)\leq d(x,y) + d(y,x)$ (which is the triangular inequality) 
I know that the metric topology is  the topology on X generated by the basis $\{B_{\epsilon,d}(x): \epsilon > 0, x \in X\}$
I know what all these things mean distinctly but I don't know how I would be able to start a proof like this. Can someone please help? 
 A: You can always try to see it geometrically in the first place, even only getting a picturesque idea, which may still help.
Here is a not very rigorous proof using the geometric intuition as a guide (because what is asked is how to start a proof).
Let $(x,y)$ lie in the unit square without boarder of center $(0,0)$; let $d := ||(x,y)||$. If $d = 0$, then we are done (surely you can find a ball that works, right?). Suppose $d > 0$. Without loss of generality, suppose the line $x=1$ is the line, among the four bordering lines, that is closest to $(x,y)$. Let $d'$ be the distance between $(x,y)$ and the line $x=1$. Then $\min (d, d') > 0$. But then the open ball of center $(x,y)$ and radius $\min (d,d')$ lies in the square, so we are done. 
A: Let $(x,y) \in \mathbb R^{2}$ and $0<x<1,0<y<1$. Let $r$ be the minimum of the four numbers $x,1-x,y,1-y$. Since all the four numbers are $>0$  we get $r>0$. Suppose $(u,v)$ is any point of $\mathbb R^{2}$ whose  distance from $(x,y)$ is less than $r$. Then verify that $0<u<1$ and $0<v<1$, proving that $\{(u(v): 0<u<1,0<v>1\}$ is open. For example, $|u-x|<r$ implies $u-x <1-x$ so $u<1$. Nexr $|u-x| <x$ gives $x-u<x$ so $u>0$. Now go ahead and give a similar argument for $v$.
A: The unit square = (0,1)×(0,1), is the product of two open sets, hence open.
