To prove $ U = \{ (x,y)^T : x>y\} \subset \mathbb{R}^2 $ is an open set Let $(x,y)^T$ be a point in $U$. 
I pictured the graph that for any point whose distance from $(x,y)$ is less than the shortest distance of $(x,y)^T$ from the line $x=y$, then the point is in $U$.
Let $(a,a)^T$ be a point on the line $x=y$ so that $(a,a)^T$ is the closet point to $(x,y)^T$ from the line.
So that $(y-a)/(x-a)=-1$ as the slope of $x=y$ is 1 and the shortest line should be perpendicular to it, which I get $a = (x+y)/2$. 
And the shortest distance of $(x,y)^T$ from the line $x=y$ should be $\sqrt{(y-\dfrac{x+y}{2})^2 + (x-\dfrac{x+y}{2})^2 }$, which is $\dfrac{x-y}{\sqrt{2}}$ .
I want to show that when $\delta=\dfrac{x-y}{\sqrt{2}}$, any point in $B_\delta(\begin{array}\\x\\y\end{array})$ is in $U$. But I have no idea how to proceed.
 A: Alternative way: it's the preimage of the open set $(0, \infty)$ under the continuous map $f: \mathbb{R}^2 \to \mathbb{R}$ given by $(x, y) \mapsto x-y$.
A: Here is a simple geometric approach. Let $(x,y)^T$ be in $U$. As you say, the closest distance to the line $y=x$ is $\delta = \dfrac{x-y}{\sqrt{2}}$. Then pick a new point $p=(x+a,y+b)^T$, for which it is $\sqrt{a^2 + b^2} < \dfrac{x-y}{\sqrt{2}}$. If we can show that each such $p$ is in $U$, then it means that we have shown that $B_{\delta}((x,y)^T)$ is in $U$ which means that $U$ is open. For $p$ it must be $x + a > y + b$ or $x-y > b - a$ to be $p \in U$. 
We have $2a^2 + 2b^2 < (x-y)^2$. It is: $$2a^2 + 2b^2 - (b-a)^2 = 2a^2 + 2b^2 -b^2 + 2ab - a^2 = (a+b)^2 \geq 0$$
Hence $(x-y)^2 > 2a^2 + 2b^2 \geq (b-a)^2$ and $|x-y| > |b-a|$.
We know that $x-y > 0$, which means, with the above result we have $x-y > b-a$.
A: Suppose $(a, b) \in U$. Then $a>b$, and so there exists $\varepsilon>0$ such that $a>b+2\varepsilon $. If $(x, y) $ has distance less than $\varepsilon$ from $(a, b) $, then $x>y$ (I'll leave this for you to show), i.e. $(x, y) \in U$. This shows $U$ is open. 
A: Hint:
Easier: Show $U^c = \{(x,y): x \leq y\}$ is closed!
Patrick Stevens' approach is also nice.
