The open $(0,1) \times (0,1)$ square invectively mapped *into* the interval $(0,1)$ Does the following bijection work: Take any point $(x,y) \in (0,1) \times (0,1).$ Each real number $r \in (0,1)$ may be represented by an infinitely-long decimal expansion (0.235, for example, is the same as 0.234999999...). Take the real numbers $x,y \in \mathbb{R}$ and interlace their decimal expansions to produce a unique real number $r' \in (0,1).$ The number $r'$ being unique, the mapping is 1-1. Given a real number $r' \in (0,1),$ one can unlace the decimal expansion of that number according to the pattern set by the mapping and produce the real numbers $x$ and $y$, and therefore arrive at a unique point $(x,y) \in (0,1) \times (0,1).$ Does a bijection exist?

Not quite. The correspondence between infinite decimals and elements of $(0,1)$ is not itself a bijection because some numbers have more than one decimal expansion.

You solve that by restricting to the expansion with infinitely many non-zero digits, fine. But now the mapping from $(0,1)^2$ to $(0,1)$ is not surjective. For example there are no $x$ and $y$ that give $r'=0.1101010101...$. Because that would require $x=0.10000...$ and $y=0.111...$. But $x=0.1000...=0.09999...$, so the image of that $(x,y)$ is actually $0.0191919...\
e0.11010101...$.