Can I please get feedback on my proof for the problem below? Thank you!
$\def\R{{\mathbb R}} \def\Rhat{{\widehat{\R}}}$
So we know the interval $(0,1)$ is open in $\R$. Prove that, when considered as a subset of $\R^2$, that is, as a line segment on the $x$-axis in the plane, it is not open. More specifically, show that the set $(0,1)\times\{0\} \subseteq\R^2$ is not open.
$\textbf{Solution:}$ Let $S = (0,1) \times \{0\}$ is open in $\R^2$. Then, by definition, if we take a ball, say $B(x, r)$ centered at any point $x \in S$ with sufficiently small radius $r>0$, then $B(x,r) \subset S$.
Now, take $x = (\frac{1}{2}, 0), r = \frac{1}{2}.$
Then the ball $\displaystyle{\mathbb{B} ((\frac{1}{2}, 0), \frac{1}{2})}$ is a circle whose center is at $(\frac{1}{2}, 0)$ and radius is $\frac{1}{2}$.
By our assumption as $S$ is open in $\R^2$ and $(\frac{1}{2}, 0)\in S$, therefore, the ball $\displaystyle{\mathbb{B} ((\frac{1}{2}, 0), \frac{1}{2})}$ should entirely lie within $S$ but observe that $\displaystyle{(0,\frac{1}{4}) \in \mathbb{B} ((\frac{1}{2}, 0), \frac{1}{2})}$ while $(\frac{1}{2}, 0)\notin S$, hence a contradiction. Thus, our assumption is not true, that is, the set $S= (0,1) \times \{0\}$ is not open in $\R$
"if we take any ball" can't be right. Balls around points in the open disk with radius $1$ can't meet this prescription -- consider any point in this disk and set the radius of the ball to $1$ million. It is the case that every point in an open set is the center of an open ball contained in the set (but its radius may have to be small, especially if the point is near the boundary of the set).
To adjust your proof to incorporate this modification, you should show that for all $r>0$, $B((\frac{1}{2},0),r)$ fails to be contained in $S$.