Locus of the fibered product of two affine lines with doubled origin where projections coincide

Question

This is an exercise (Exercise III-1) in the book Geometry of Schemes:
Exercise III-1. a) Let $Y$ be the line with doubled origin over a field $K$ and let $\varphi_1,\varphi_2:\mathbb{A}_K^1\to Y$ be the two inclusions. Show that the locus where $\varphi_1$ and $\varphi_2$ agree (as continuous maps of topological spaces) is not closed.
b) Let $X:=Y\times_K Y$ and let $\varphi,\psi$ be the two projection maps from $X$ to $Y$. Show that the set of points at which $\varphi$ and $\psi$ agree is not closed. Show that the same is true for the set of closed points at which the projections agree.
My approach. Exercise a) is not that difficult. The locus is exactly $D(t)\subseteq \mathbb{A}_K^1$, which is open.
b) Since the fibered product $X$ is constructed by gluing affine pieces that cover it. I tried to show that such locus on each affine piece is not closed. As $Y$ is covered by $Y_1\cong \operatorname{Spec}k[x]$ and $Y_2\cong \operatorname{Spec}k[y]$, the product $X$ should be covered by four affine pieces $Y_i\times_K Y_j$ where $i,j=1,2$. Now each $Y_i\times_K Y_j\cong \mathbb{A}_K^2$ and here I encounter the problem:
Question. I want to describe those points of $\mathbb{A}_K^2$ such that they are sent to the same prime ideal, i.e. $\mathfrak{p}\in \mathbb{A}_K^2$ such that $$i_x^{-1}(\mathfrak{p})=i_y^{-1}(\mathfrak{p})$$ where $i_x,i_y$ are $k[x],k[y]\hookrightarrow k[x,y]$ corresponding to two morphisms $\mathbb{A}_K^2\to \mathbb{A}_K^1$. Isn't the possible prime only $(0)\subseteq k[x,y]$? In that case, on each affine cover $Y_i\times Y_j$, the locus is the generic point, which is open. Hence the locus on the whole $X$ where $\varphi,\psi$ agree is also not closed. But it gets weird when considering the second minor question: the set of closed points. Since now it is basically empty.
Again, any help is sincerely appreciated!