A closed set in $\mathbb A^2_k\times\mathbb P^1_k$ Let $k$ be an algebraically closed field and consider the Zariski topology on $\mathbb A^2_k$ and on $\mathbb P^1_k$.
If
$$X:=\left\{((x_0,x_1),(y_0:y_1))\in\mathbb A^2_k\times \mathbb P^1_k\,\bigg| x_0y_1=x_1y_0\,\right\}\subseteq \mathbb A^2_k\times \mathbb P^1_k$$
I don't understand why $X$ is closed in $\mathbb A^2_k\times \mathbb P^1_k$.
The structure of variety on $\mathbb A^2_k\times \mathbb P^1_k$ is obtained by patching of the affine varieties $\{\mathbb A^2\times U_i\}_{i=0,1}$ where $U_1=\{(x_0:1)\,:x_0\in k\}$ and $U_0=\{(1:x_1)\,:x_1\in k\}$. Now is there some kind of "characterization" for closed sets of $\mathbb A^2_k\times \mathbb P^1_k$?
 A: Yes, there is a characterization of closed subsets of $\mathbb A^2_k\times \mathbb P^1_k$.  
Namely, they are the common zeros of a family $(f_i(x_0,x_1;y_o,y_1))_{i\in I}$ of polynomials $f_i(x_0,x_1;y_o,y_1)$, required to be homogeneous in the variables $y_0,y_1$ associated to $\mathbb P^1_k$.
The degree of homogeneity of these polynomials may vary with $i\in I$.
Your example is a little confusing (by no fault of yours!) because the polynomial $x_0y_1-x_1y_0$ happens to be homogeneous (of degree $1$) also in the variables $x_0,x_1$ of $\mathbb A^2_k$, which is absolutely not required in general.
This means that your subvariety $X\subset \mathbb A^2_k\times \mathbb P^1_k$ descends to a subvariety $\mathbb P(X)\subset \mathbb P^1_k \times \mathbb P^1_k$.    
A geometric interpretation
You have the Plücker embedding $$p:\mathbb P^1_k \times \mathbb P^1_k\hookrightarrow \mathbb P^3_k: ((x_0:x_1),(y_0:y_1))\mapsto (z_0:z_1:z_2:z_3)=(x_0y_0:x_0y_1:x_1y_0:x_1y_1)$$The image $p(\mathbb P(X))\subset \mathbb P^3_k$ of $\mathbb P(X)$ under that embedding is then the intersection of the quadric $z_0z_3-z_1z_2=0$ with the plane $z_1-z_2=0$.
 In other words, a good old projective conic! 
