$A_r=\{(x, y) \in \mathbb{R} \times \mathbb{R} : x+y=r\}$ I have figured out how to prove $A_r$ is nonempty but I am stuck on the last two parts of proving a partition of Real numbers. For part 2 I have 
"Let r, s $\in \mathbb{R}$ with x+y=r and x+y=s, so there exists $A_r$ and $A_s$."
 A: I'm assuming you are trying to show that $\{A_r\}_{r \in \Bbb{R}}$ forms a partition of $\Bbb{R}^2$.
Clearly for each $r \in \Bbb{R}$, the point $(r,0) \in A_r$, hence $A_r \neq \emptyset$.
To prove: $\Bbb{R}^2=\bigcup_{r \in \Bbb{R}}A_r$.
Let $(a,b) \in \Bbb{R}^2$, then consider the real number $r_0=a+b$. Corresponding to this $r_0$ we have the set $A_{r_0}$. Now $(a,b) \in A_{r_0} \subseteq \bigcup_{r \in \Bbb{R}}A_r$. Thus $\Bbb{R}^2 \subseteq \bigcup_{r \in \Bbb{R}}A_r$. The other containment is obvious.
To prove: $A_r \cap A_s = \emptyset \iff r \neq s$.
Assume $r \neq s$ and suppose $(c,d) \in A_r \cap A_s$. Then $(c,d) \in A_r$, which means $c+d=r$. Also $(c,d) \in A_s$, which means $c+d=s$. But this contradicts the fact that $r \neq s$. So the intersection is empty. The other implication is straightforward.
A: Consider the line in the Cartesian plane joining $(1,0)$ and $(0,1)$. This set of points of this line is your $A_1$. 
Now consider ALL the lines in the plane that are parallel to this $A_1$. As parallel  lines do no intersect they are disjoint. 
And also given any point in the plane not in the line $A_1$ there is a (unique) line passing through that and parallel to $A_1$. So every point is covered by these lines. 
Any such line is not parallel to the $x$-axis and so it will intersect it at some point $(r,0)$, and it is the line $A_r$ in your problem. Thus you can see that these sets provide a partition.
