Showing the subset $\{(x_1,x_2) \in \mathbb{R}^2 : x_1 > x_2 \}$ is open The metric is the typical Euclidean metric, $ \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } $.
I have solved this one, albeit with in my opinion quite excessive steps. I would love to know if there is a simpler way to do it. Below is my approach.

Denote $A = \{(x_1,x_2) \in \mathbb{R}^2 : x_1 > x_2 \}$.
Take $p \in A$ and write $p = (x_1, x_2)$, where $x_1 > x_2$. Want to find a ball around $p$ entirely contained in $A$.
The smallest distance from $p$ to the line $x_2 = x_1$ is $\frac{x_1-x_2}{\sqrt 2}$.
Take any smaller radius, such as $ r = \frac{x_1 - x_2}{10} > 0$. Want to show the open ball $B_r (x) \subset A$.
Let $q \in B_r (x)$ and write $ q = (y_1, y_2)$. Want to show $y_1 > y_2$.
Since $q \in B_r (x)$, we have $\sqrt{ (x_1 - y_1)^2 + (x_2 - y_2)^2 } < r$.
This implies respectively
$$|x_1 - y_1| < r \ \ \ \text{  and  } \ \ \ |x_2 - y_2| < r,$$
owing to $|a| \leqslant \sqrt{a^2 + b^2} < r$.
Adding these inequalities we get the condition $ |x_1 - y_1| + |x_2 - y_2| < 2r $.
Using the symmetry of absolute value, write $|x_1 - y_1| = |y_1 - x_1|$. The triangle inequality in reverse gives us:
$$|y_1 - x_1 + x_2 - y_2| \leqslant |x_1 - y_1| + |x_2 - y_2| < 2r$$
Exploiting the fact that $2r = \frac{x_1 - x_2}{5} < x_1  - x_2 $, we now write: $$ x_2 - x_1 < -2r < y_1 - x_1 + x_2 - y_2 < 2r < x_1 - x_2 $$
Adding and subtracting $x_1$ and $x_2$ respectively, we get:
$$ 0 < y_1 - y_2 < 2(x_1 - x_2) $$
Since $x_1 > x_2$, this inequality is legitimate and shows $y_1 > y_2$. Done.

Could there by a simpler way to show this? I found all the inequality juggling somewhat circuitous, which made me wonder whether there is a slightly more shorter and elegant way to show it.
 A: You can define $f:\mathbb{R^2}\to\mathbb{R}$ by $f(x,y)=x-y$. This is a continuous function, and $A=f^{-1}((0,\infty))$, the inverse image of an open set. Hence $A$ is open.
A: The simplest way is as follows: consider the map $f\colon\Bbb R^2\longrightarrow\Bbb R$ defined by $f(x_1,x_2)=x_1-x_2$. Then $A=f^{-1}\bigl((0,\infty)\bigr)$. So, since $f$ is continuous and $(0,\infty)$ is open, $A$ is open.
A: You are off to a good start. Here is a shorter version of your argument: let $p \in A$ and write $p = (x_1,x_2)$ so of course $x_1 > x_2$.  Define $r = x_1 - x_2$.
Select $0 < \epsilon < \dfrac r2$. If you are given a second point $q = (y_1,y_2)$ with $|p-q| < \epsilon$ then
$$|x_1 - y_1| \le |p-q| < \epsilon$$
and
$$|x_2 - y_2| \le |p-q| < \epsilon$$
so that
$$y_2 \le x_2 + \epsilon = x_1 - r + \epsilon \le y_1 - r + 2\epsilon < y_1$$
and in particular, $q \in A$.  It follows that $B(p,\epsilon) \subset A$.
This implies every $p \in A$ has a neighborhood contained in $A$ so that $A$ is open.
A: Let $$ A := \{(x_1,x_2) \in \mathbb{R}^2 : x_1 < x_2 \},$$
then the complement of $A$ is
$$A^c = \{(x_1,x_2) \in \mathbb{R}^2 : x_1 \geq x_2 \}.$$ Showing that $A^c$ closed will show us that $A$ is open.
Let $(x_k)_{k\in\mathbb{N}} \subseteq A^c$ be such that $x_k \xrightarrow{d} x $ to some $x \in \mathbb{R}^2$ as $k \to \infty$, and represent $x_k = (x_{1,k},x_{2,k})$ for each $k \in \mathbb{N}$. Since $$x_{1} = \lim_{k\to\infty} x_{1,k} \geq \lim_{k\to\infty} x_{2,k} = x_2,$$ we must have that $x_1 \geq x_2$ and thus $x \in A^c$. So $A^c$ is closed, and thus $A$ is open.

Note: $d$ is the Euclidean metric.

