Proving set $(0,1)\times\{0\} \subseteq\mathbb{R^2}$ is not open. Can I please get some help/feedback on my proof? Thank you.
$\def\R{{\mathbb R}}                            
\def\Rhat{{\widehat{\R}}}                  
\def\N{{\mathbb N}}$ 
$(0,1)$ is open in $\R$.  I will 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. Specifically, I will show that the set $(0,1)\times\{0\} \subseteq\R^2$ is not open.
$\textbf{Solution:}$ Consider the set $(0,1) \times \{0\} \subset \R^2.$ A set $S$ is called open if every point of the set $S$ is an interior point, that is, for every point $x\in S$, there exist an open set $V$, such that $x\in V \subset S.$
Pick the point $p = (\frac{1}{2}, 0)$ inside $(0,1) \times \{0\}$. In $\R^2$, the open balls form a basis for the topology of $\R^2$, meaning every point $x$ in $\R^2$, we can find an open ball containing it and if an open set $U$ contains $x$, there exists an open ball centered at $x$ such that $x\in B \subset U$. So, if we can show there do not exist any open ball centered at $p = (\frac{1}{2}, 0)$ contained in $(0,1) \times \{0\}$, we will be done. 
Now, we will show why no open ball sits inside $(0,1) \times \{0\}$ by supposing it is, that is, there is an open ball $B(p,r)$, for some $r>0$,  in the Euclidean metric on $\Bbb R^2$ such that $$B(p,r) \subseteq (0,1) \times \{0\}\tag{1}$$
But $q=(\frac12, \frac{r}{2})$ obeys $d(p,q)=\frac{r}{2}< r$, so that $q \in B(p,r)$ but as $\frac{r}{2} \neq 0$, $q \notin (0,1) \times \{0\}$. This contradicts our supposed inclusion $(1)$. So $p$ is not an interior point of $(0,1) \times \{0\}$ and $(0,1) \times \{0\}$ is not open.
 A: The idea is fine: show that e.g. $p=(\frac12,0)$ is not an interior point of $(0,1) \times \{0\}$. But you do not show why no open ball sits inside $(0,1) \times \{0\}$, you need to fill that gap (e.g. a picture is not a proof!)
So suppose it is, so there is an open ball $B(p,r)$, for some $r>0$,  in the Euclidean metric on $\Bbb R^2$ such that $$B(p,r) \subseteq (0,1) \times \{0\}\tag{1}$$
But $q=(\frac12, \frac{r}{2})$ obeys $d(p,q)=\frac{r}{2}< r$, so that $q \in B(p,r)$ but as $\frac{r}{2} \neq 0$, $q \notin (0,1) \times \{0\}$. This contradicts our supposed inclusion $(1)$. So $p$ is not an interior point of $(0,1) \times \{0\}$ and $(0,1) \times \{0\}$ is not open.
A: Another idea: take for example the sequence 
$$\left\{\left(\frac12\,,\,\frac1n\right)\right\}_{n\in\Bbb N}$$
Check that the above sequence is not in the set $\;E:=(0,1)\times\{0\}\;$, so it belongs to $\;E^c=\Bbb R^2\setminus E\;$ . The sequence is clearly convergent, so if $\;E^c\;$ is closed it must contain this sequence's limit, yet
$$\lim_{n\to\infty}\left(\frac12\,,\,\frac1n\right)=\left(\frac12,\,0\right)\in E$$
which means $\;E^c\;$ is not closed and thus $\;(E^c)^c=E\;$ is not open. $\;\;\;\;\;\;\blacksquare\;$
