Definition. Boundary Points. Let $A\subset \mathbb{R}^n$. A point $x\in \mathbb{R}^n$ is called a boundary point of A if every neighborhood of $x$ contains at least one point in $A$ and atleast one point not in $A$.
Let $ S $ be the set $ \mathbb {R} ^ 2 $ with the origin removed, show that $ \vec{0} $ is a boundary point of $ S $.
Proof. To show that $ \vec{0} = (0, \: 0) $ we have to show that for all $ \varepsilon> 0 $ the following is true for the interval $ I = (0- \varepsilon, \: 0 + \varepsilon) $:
$$ I \cap (0, \: 0) \neq \emptyset \wedge I \cap S \neq \emptyset $$
Clearly, $ 0 - \frac{\epsilon} {2} <0 $, since $ \epsilon> 0 $ and $ 0- \frac{\epsilon}{2}> 0- \epsilon $. This means:
$$0- \frac{\epsilon}{2} \in I \wedge 0- \frac{\epsilon}{2} \notin (0, \: 0)$$
Similarly, for $ x = 0 + \frac {\varepsilon} {2} $ we have
$$x \in I,\: x \in (0, \: 0)$$
This proves that $ (0, \: 0) $ is a boundary point of $ S $
It's okay? Any suggestion?
Well, while it's a nice try, you seem to have assumed, incorrectly, that a neighborhood of the origin is, or contains, an interval in $\mathbb R^{\color {red}{1}} $.
So, instead you should show that any ball in $\mathbb R^{\color {blue}{2}} $ containing $\vec0$ intersects $\mathbb R^2\setminus\{\vec0\} $.
You're off by one dimension, and should point out how you pass to an interval.