Does $(\Bbb{R}, +)$ admit an irreducible $2$-traversal? For a given natural number $k$, I'm going to call a subset $T$ of the plane $\Bbb{R}^2$ a $k$-traversal if, for any $x \in \Bbb{R}$,
\begin{align*}
k &= \operatorname{card} \{(a, b) \in T : a = x\} \\
&= \operatorname{card} \{(a, b) \in T : b = x\} \\
&= \operatorname{card} \{(a, b) \in T : a + b = x\}.
\end{align*}
It's not difficult to see that lines in the plane that aren't parallel to the $x$-axis, the $y$-axis, or the line $x + y = 0$ will be $1$-traversals (but are far from the only examples!). We can make $2$-traversals without difficulty by taking two disjoint $1$-traversals and unioning them. My question is,

Is there a $2$-traversal that cannot be decomposed into the union of two $1$-traversals?

The idea of (irreducible) $k$-traversals is a concept from latin squares. I'm trying to consider the concept when applied to an (uncountably) infinite Latin square generated by the group $(\Bbb{R}, + )$.
This was a question that popped into my head years ago when attending a combinatorics conference. Combinatorics is not my forte, but it seemed like an interesting question, and I thought I'd share it.
 A: Vincent's example... $T$ is the union of two graphs $y=\frac{|x-1|}{2}$ and $y=-\frac{|x+1|}{2}$.

Each vertical line meets $T$ in two points, one in the top graph and one in the bottom graph.
Each horizontal line meets $T$ in two points: either two in the top graph, or two in the bottom graph, or the two corners on the $x$-axis.
Each line of slope $-1$ meets $T$in two points, one in the top graph and one in the bottom graph.
Thus, $T$ is a $2$-traversal.
Next, we claim it is impossible to have $T = A \cup B$ where $A$ and $B$ are $1$-traversals and $A \cap B = \varnothing$.  Indeed, consider the three points
$$
u:=\left(0,\frac12\right),\quad
v:=\left(2,\frac12\right),\quad
w:=\left(2,-\frac32\right).
$$

Now $A$ contains exactly one of the points $u,v$, since they are both on the line $y=\frac12$.  $A$ contains exactly one of the points $v,w$, since they are both on the line $x=2$.  And $A$ contains exactly one of the points $w,u$, since they are both on the line $x+y=\frac12$.  Impossible.
A: I have the feeling that this (or a variation) would work, but maybe I am overlooking something. It is definitely the union of two things, but they are not 1-traversals.
Take the union of the v-shape $y = |1/2(x-1)|$ and upside-down v-shape $y = -|1/2(x+1)|$.
