After thinking some time on this question I found a relatively simple solution based on the well-known fact that all arcs in the plane are ambiently homeomorphic. Using this fact and assuming that an embedding $f:A\to\mathbb R^2$ exists, we can modify $f$ and assume that $f(0,t,0)=(0,t)$ for all $t\in[-1,1]$. By the continuity of $f$, find $\epsilon>0$ such that $f(x,y,z)\in \mathbb R\times(-\frac12,\frac12)$ for any point $(x,y,z)\in A\cap [-\epsilon,\epsilon]^3$. Then use the connectedness argument to prove that the images of the sets $A_-=A\cap([-\epsilon,0)\times[-\epsilon,\epsilon]\times\{0\})$ and $A_+=(A\cap(0,\epsilon]\times[-\epsilon,\epsilon]\times\{0\})$ are containined in distinct connected components of $(\mathbb R\setminus \{0\})\times[-\frac12,\frac12]$. One of this components should contain also the connected set $f(\{(0,0)\}\times(0,\epsilon])$. Finally, applying a known lemma from Dimension Theorem (saying that two arcs linking opposite sides of a square have a common point) we can prove that the arc $f(\{(0,0)\}\times[0,\epsilon])$ has non-empty intersection with an arc $f(\{(\delta,0)\}\times[-1,1])$ for some small non-zero number $\delta$. But this contradicts the injectivity of $f$. This contradition completes the proof of the non-planarity of $A$.

So, sorry for boring you with this (relatively simple) question. (But the solution was not immediately evident).