If you assume the axiom of choice it's enough to have surjections both way.
For the most reasonable attempt at flooding a plane with R, I still end up with something where I can trivially define values {r0, r1} that do not map back to R.
x=0.a1a2a3a4a5a6...
(always pick the nonrepeating decimal expansion in case x has two) and sending it to the pair (y,z) where
y=0.a1a3a5... z=0.a2a4a6...
(Using (0,1) is no issue since we can compose later with bijections (0,1) -> R)
Edit: I'm not sure why I cannot reply to your reply, but {π,π} is not even in (0,1)^2
We can also do a map f:R -> R^2 directly, it's just slightly more annoying to write. Take a real number x, written in decimal expansion as
x=bn...b2b1b0.a0a1a2a3a4...
where again we choose the representation which is not repeating in case x has two. We also choose our first index n so that bn is zero but b(n-1) is not. We now map x to the pair (y,z), where
y=...b5b3b1.a1a3a5...
z=...b4b2b0.a0a2a4...
So that, for example, (π,π) is the image of x=33.114411559922...
Sorry, didn't even look at the bounds - assumed things went to infinity.