Showing the diagonal on $\mathbb{R}^2$ is closed $D=\{{(x,x)|x\in\mathbb{R}}\}$ , I want to show D is closed with the definition: 

$A$ is closed if and only if $\mathbb{R}/A$ is open.

So basically what I really want to show is $A=\{{(x,y)\in \mathbb{R}^2|x\neq y\}}$
.
My attemp so far was taking $a=(x,y)\in A$ and an open ball $B(a,r)$ with $r=\frac {|x-y|} {2\sqrt 2}$, which is the distance between $a$ and the line $x-y=0$, divided by two.
It is easy to see geometrically why a ball with this radius is contained in A, but I wasn`t able to complete the proof formally, using algebra.
 A: Let's try to replicate the proof for the general statement I linked to in the comments to this context, making it more elementary.
So, you want to prove that $A=\{(x,y) \in \mathbb{R}^2 \mid x \neq y\}$ is open. Given two different $x,y \in \mathbb{R}$, we can separate them by two intervals $(a,b)$ and $(c,d)$. Specifically, $(x-|x-y|/2,x+|x-y|/2)$ and $(y-|x-y|/2,y+|x-y|/2$) are disjoint neighbourhoods of $x$ and $y$, respectively. If you have seen metric spaces you will realize that they are just the open balls of half the distance between one point and the other (and this is essentially just the proof that metric spaces are what is called a Hausdorff space: different points can be separated by neighbourhoods). The fact that they are disjoint is easily seen from the triangle inequality.
Now, call those two neighbourhoods $V_x$ and $V_y$. Then, $V_x \times V_y$ does not intersect the diagonal, and thus is still contained in $A$. This is clear from the fact that for a set of the form $E \times F$, intersecting the diagonal is equivalent to having a point in $E \cap F$.
Note that $V_x \times V_y$ is a rectangle around the point $(x,y)$ (which is on its center). If you pick the smallest of the diameters of the sets $V_x,V_y$, then the open ball with radius of half this amount and center on $(x,y)$ is clearly contained in $V_x \times V_y$, and thus in $A$ (if this is not clear, prove it. It is essentially the inequality $\Vert x\Vert_{\infty} \leq \Vert x \Vert_2$). Since $(x,y) \in A$ was arbitrary from the start, this means that $A$ is open.
This proof has some advantages. First, it is related intrinsically to the proof of the general statement: merely purging away the context yields a complete proof. It also implicitly refers to the fact that the product topology and the norm topology are equivalent (more precisely, the argument actually goes through the fact that the norm topology is finer than the product topology).
A: Your approach is fine. In fact (as you can see geometrically), there's no need to divide by $2$. In other words, you can take $r=\frac{|x-y|}{\sqrt2}$. I will prove that the open disk $D\bigl((x,y),r\bigr)$ contains no element of $D$, that is, I will prove that if $z\in\mathbb R$, then $\bigl\|(x,y)-(z,z)\bigr\|\geqslant r$. In fact,\begin{align}\bigl\|(x,y)-(z,z)\bigr\|<r&\iff(x-z)^2+(y-z)^2<\frac{(x-y)^2}2\\&\iff2\bigl((x-z)^2+(y-z)^2\bigr)<(x-y)^2\\&\iff x^2+y^2+4z^2+2xy-4xz-4yz<0\\&\iff(x+y-2z)^2<0,\end{align}which is impossible, of course.
A: The following applies:
A topological space $X$ is a Hausdorff space if and only if its diagonal 
$$\Delta X = D = \{(x,x)\mid x \in X\} \subset X \times X$$ 
is closed in $X\times X$
Since we have $X = \mathbb{R}$, we can show that $D = \{(x,x)\mid x \in \mathbb{R}\}$ is closed in $\mathbb{R}^2$.
Let $(x,y) \in \mathbb{R}^2\setminus D $ such that $x \not= y$. Since $\mathbb{R}$ is a Hausdorff-Space there are disjoint neighbourhoods $U$ for $x$ and $V$ for $y$. Therefore $U\times V$ is an open neighborhood of $(x,y) \in \mathbb{R}^2$. It holds that $(U\times V)\cap D = \emptyset$. 
Since $\mathbb{R}^2\setminus D$ is a union of such open neighbourhoods, it holds that $ \mathbb{R}^2\setminus D $ is open. Hence $ D $ is closed in $\mathbb{R}^2$ 
A: If $(a,b)\in \mathbb{R}^2\setminus D$, then $a\neq b$.
The value $d=\frac{|b-a|}{\sqrt{2}}$ is the distance from $(a,b)$ to the diagonal. Since $a\neq b$, then $d>0$. Therefore, the open ball $B_{d/2}(a,b)=\{(x,y)\in\mathbb{R}^2:\ \sqrt{(x-a)^2+(y-b)^2}<d/2\}$ has $(a,b)$ as its center and doesn't contain points of $D$.
A: You could use Cauchy-Schwarz: $$\sqrt{2} \lVert (x,y) - (z,z) \rVert = \lVert (1, -1) \rVert \, \lVert (x,y) - (z,z) \rVert \ge | \langle (1, -1), (x, y) - (z, z) \rangle | = | x-y |. $$
Therefore,
$$\lVert (x, y) - (z, z) \rVert \ge \frac{|x-y|}{\sqrt{2}}.$$
As a consequence, if $\lVert (x, y) - (x', y') \rVert < \frac{|x-y|}{\sqrt{2}}$, then it is impossible to have $(x', y') = (z, z)$ for any $z$, hence $(x', y') \in \mathbb{R}^2 \setminus D$.
A: Parameterize the diagonal as $t\mapsto(t,t)$.
The distance between $a$ and $(t,t)$ is $\sqrt{(x-t)^2+(y-t)^2}$
The square of this distance is a quadratic function of $t$. Show that it is always greater than the square of your $r$.
A: Note that the distance between a point $X = (x,y)$ and the line $x-y=0$ is $\frac{|x-y|}{\sqrt{2}}$.
Now, consider a point $X = (x,y)$ with $x\neq y$ (i.e., $X \in A $),  and a ball $B_r(X,r)$ centered at $X$ with $r = \frac{|x-y|}{2 \sqrt{2}}$.
Consider any point $C = (c,c)$ (i.e, $C\notin A$).
$d(C,X) = \sqrt{(x-c)^2+(y-c)^2} = \sqrt{x^2 + y^2 + 2c^2 -2cx-2cy}$
Now let us find $C = (c,c)$ for which $d(C,X)$ achieves the minimum value.
Let $f(c) = x^2 + y^2 + 2c^2 -2cx-2cy$
$ f'(c) = -2x - 2y + 4c $
$f'(c) = 0 \implies c = \frac{x+y}{2}$
$ C_{min} = (\frac{x+y}{2},\frac{x+y}{2})$
So minimum of $d(C,X)$ for $C \notin A $is  $d(C_{min},X) = \frac{|x-y|}{\sqrt{2}} \ge r$
So, $C \notin A \implies C \notin B_r(X,r)$ 
