Homeomorphism between a subset of $\mathbb{R}^3$ and $\mathbb{R}^2$.

I consider the set, for $\varepsilon>0$ fixed :

$$E=\{(x,y,z)\in\mathbb{R}^3 :x\in\mathbb{R},z\in\mathbb{R}, \exists d\in(0,\varepsilon), y= d-z\}$$

We first observe that we have two free variables, let’s say it is $x$ and $z$. From this observation and some (misleading maybe) drawing, I would like to prove that the set is homeomorphic to $\mathbb{R}^2$.

Seeking for an homeomorphism encoding all the information, I first thought to try $f(x,y,z)=(x,d)\in\mathbb{R}\times(0,\varepsilon)$. However this is clearly not injective.

Thus I tried to exploit the free variable and looked at $f(x,y,z)=(x,z)\in\mathbb{R}^2$ however this does not work, especially because $d$ varies on $(0,\varepsilon)$.

The case where $d$ is fixed is easier, but I do not see how to recover the set $E$ starting from this.

If you have some hints to provide, I would be happy to learn it !

Thank you a lot