Why does every neighbourhood contain this point? First I have the K-topology $\mathbb{R}_K$ Then I create the quotient space Y, by collapsing K to one point. That is, for each $x \in \mathbb{R}_k$ we have a function p, and if x is not in K, $p(x)=x$, if x is in K, we send x to the point representing K. Then we give Y the quotient topology. That is, a set is U open in Y iff $p^{-1}(U)$ is open in $\mathbb{R}_K$.
In a solution to an exercise, it is stated that the diagonal on $Y \times Y$ is not closed. Because every neighborhood around $(K,0)$, contains $(0,0)$. But why is this?
Please look at where I argue this is not the case, is my argument wrong? The set $(0,2)$ is open in $\mathbb{R}_K$, it is also a saturated set, beacuse it contains K, then $p((0,2))$ is open in Y. This set contains the point K in y, but not zero. Then in $Y\times Y$, $(0,2)\times (-2,2)$(here (0,2) and (-2,2), are not intervals, but the point K, + the other points in the interval) is an open set in $Y\times Y$, containing $(K,0)$, that does not contain $(0,0)$. Is this correct?
For reference, the exercise this comes up in is part b here:
http://s28.postimg.org/ysamndbp7/image.png
And the solution I am talking about is here:
http://dbfin.com/topology/munkres/chapter-2/section-22-the-quotient-topology/problem-6-solution/
But if you look at this links, you will see that what I am asking about, is just a small part, of a much bigger problem, and it is only the part I have asked about above that I am asking about here, not the entire question.
 A: Corrected. Let $G$ be an open nbhd of $\langle K,0\rangle$ in $Y\times Y$. Without loss of generality we may assume that $G=W\times U$ for some open $U$ and $W$ in $Y$. Let $V=p^{-1}[U]$; $V$ is an open nbhd of $0$ in $\Bbb R_K$, so there is an $\epsilon>0$ such that $(-\epsilon,\epsilon)\setminus K\subseteq V$. Let $H=p^{-1}[W]$; $H$ is an open nbhd of the set $K$ in $\Bbb R_K$. 
Choose $n\in\Bbb Z^+$ big enough so that $\frac1n<\epsilon$; then the open interval $\left(\frac1{n+1},\frac1n\right)$ is a subset of $V$. Moreover, there is a $\delta>0$ such that $\left(\frac1n-\delta,\frac1n+\delta\right)\subseteq H$ and $\frac1n-\delta\ge\frac1{n+1}$. Let $x\in\left(\frac1n-\delta,\frac1n\right)$, then
$$x\in\left(\frac1n-\delta,\frac1n\right)\cap\left(\frac1{n+1},\frac1n\right)\subseteq H\cap V\;,$$
so $\langle p(x),p(x)\rangle=\langle x,x\rangle\in p[H]\times p[V]=G$. Thus, every open nbhd of $\langle K,0\rangle$ in $Y\times Y$ contains a point of the diagonal, which is therefore not closed.
(You are quite right in thinking that $\langle K,0\rangle$ has nbhds that do not contain $\langle 0,0\rangle$; the result is correct, but the argument has to be a bit more complicated.)
