Let $U\subseteq\mathbb{R}^2$ be defined as $\{$ $(x_1,x_2) \in \Bbb R^2$ $:$ $x_1>0$ $\}$. Then $U$ is open.
My proof:
Let $\textbf{x} \in U$ then $\textbf{x}= (x,y)$ where $x>0$. Let $r= x$. If $\textbf{y}$ $=$ $(y_1,y_2)$ and $\textbf{y}\in B(\textbf{x},r)$ then
$\|\textbf{x}-\textbf{y}\|$ $=$ $\sqrt{(x-y_1)^2+(y-y_2)^2}$ $<x$ $\implies$ $(x-y_1)^2<x^2$ and so either of the following hold:
$(1)$ $x-y_1-x<0$ and $ x-y_1+x>0$. This implies that $y_1>0$.
(2) $x-y_1-x>0$ and $x-y_1+x<0$. This implies that $x<\frac{y_1}{2}<0$, a contradiction, so (1) must hold and thus $\textbf{y} \in U$.
Please answer if what I have written is correct or not and then provide feedback, please.
Overall, what you did is correct. However, I would suggest to (1) provide a better explanation of the overall proof and (2) simplify the technical details.
I would do something like this:
In order to proof that $U$ is open, I have to prove that for any $x=(x_1,x_2) \in U$, $U$ contains an open disk centered on $x$. So let's pick $x=(x_1,x_2) \in U$ and prove that that the open disk $B(x,x_1)$ is included in $U$. By definition if $y=(y_1,y_2) \in B(x,x_1)$, we have $$\vert x_1 - y_1 \vert \le\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} < x_1$$ and therefore $$-x_1 < x_1-y_1 < x_1$$ and $y_1 >0$ as desired to prove that $y \in U$.