Prove $\{ (x,y) \in \mathbb{R}^2 \mid x \ne y\} \cup \{(0,0)\} $ is not open I'm trying to solve the following problem. 

Let $A = \{(x,y) \in \mathbb{R}^2 \mid x \ne y \} \cup \{(0,0)\}$. The
  intersection of $A$ with any horizontal or vertical line is open in
  this line, but $A$ is not an open subset of $\mathbb{R}^2$.

My attempt:
I made a drawing to make things more clear.

For the first part it seems obvious that any intersection of $A$ with a vertical\horizontal line $y = c, x = c$ is an open set on the line, since the intersection with $y = x$ will result into two open intervals:
$$ (-\infty, x) \cup (x,\infty) \quad \lor \quad (-\infty, y) \cup (y,\infty)$$
with exception for the lines $x = 0, y = 0$, which are also open in themselves.
Now for the second I've used the fact that if $A$ is open then $A = int(A)$. If $A$ was really an open set, then there would be an open ball $B_r(0,0) \subset A$, but that doesn't seems to be case since there's always a point from $y = x$ near the origin.
The problem:
I'm not a mathematician so some things that may seem obvious to me might need some work to sound\be correct. Any hints\suggestions\corrections that might improve this solution?
 A: if $A$ is open then its complement is closed. the complement of $A$ is $A^c=\{(x,y)|x=y\} \setminus \{(0,0)\}$ 
if $S=\{a_n\}$ is any sequence of nonzero real numbers which converges to zero then the sequence $\{(a_n,a_n)\}$ lies in $A^c$ but approaches the limit $(0,0)$, which does not lie in $A^c$.
hence $A^c$ is not closed. and $A$ is not open
A: A set is open if every point of the set has a neighborhood which is in the interior of the set. 
The origin belongs to the set $A$. But every open set containing the origin contains a ball of some radius, and this ball contains points from the line $y=x$, thus, there is no neighborhood of the origin contained entirely in $A$, and so $A$ is not open.
A: The idea is essentially correct. You could formally show the last sentence (instead of appealing to some picture, thought the intuition is quite valid): Suppose that $B_r(0,0) \subseteq A$. Then $(p,p) = (\frac{r}{2}, \frac{r}{2}) \in B_r(0,0) \text{ as } d((0,0), (p,p)) = \sqrt{\frac{r^2}{4} + \frac{r^2}{4}} < r$ but $(p,p) \notin A$ by definition of $A$, as $p \neq 0$.
