Let $y, z, a \in \mathbb{R}$. Is it true, that the solutions $x \in \mathbb{R}$ of the inequality $$ \lvert x - y \rvert - \lvert x - z \rvert \leq a $$ form a convex set (i.e. an interval)? I don't even know if this is true but I have not found a counterexample yet.
I tried to prove this by choosing solutions $x, \tilde{x} \in \mathbb{R}$ and verifying that that $x_t := tx + (1-t)\tilde{x}$ is a solution for $t \in [0, 1]$, too. The latter gave me problems: I tried to add $0 = t\lvert x-y \rvert - t \lvert x -y \rvert + (1-t) \lvert \tilde{x} - z \rvert - (1-t) \lvert \tilde{x} - z \rvert$ and use the reverse triangle inequality but that did not get me anywhere.
Hint
If $y=z$, then the set $J$ of the inequality solutions is either the emptyset or $\mathbb R$ depending on the sign of $a$. Hence is an interval in that case.
Without loss of generality, you can suppose $y\lt z$ for the rest of the analysis.
Then deal with the three cases $x \le y\lt z$, $y <x \le z$ and $x >z$ to get rid of the absolute values. For each cases, the solution set is the intersection of two intervals. Hence is an interval.
This enables to graph the map $x \mapsto \vert x-y \vert - \vert x-z \vert$ which is constant, then linearly growing then constant. Hence the level sets of this map are intervals or the emptyset.