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.

[![My drawing](https://i.stack.imgur.com/dPh5u.png)](https://i.stack.imgur.com/dPh5u.png)

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?