(newbie here)
I'm trying to prove that any open real interval is uncountable.
I try to proove this with a bijective function (a,b) --> R. Is there a further criteria for this than having a bijective function? If not, would it be possible to simply use an easy function like y=x, where x is an element in the real open interval (a,b) and y is an element of R.
On
If you have only proved that $\Bbb R$ is uncountable and want to prove that any interval $(a,b)$ is uncountable you can use any bijection between $(a,b)$ and $\Bbb R$. $y=x$ does not supply that bijection because all the elements outside $(a,b)$ do not have an image. The easiest way is to find a function that stretches the interval and any such function will do.
Often these are done by a chain of bijections. You can biject any open interval with any other open interval using a linear function, so instead of the general $(a,b)$ you can use an interval that is convenient, like $(0,1)$. One common one is to use the interval $(-\frac \pi 2, \frac \pi 2)$ and the function $y=\tan x$
Note that what Cantor's diagonal argument shows directly is not that $\mathbb R$ itself is uncountable, but that the particular open interval $(0,1)$ is uncountable.
Since $(0,1)\subseteq \mathbb R$, this immediately implies that $\mathbb R$ is also uncountable, but for your purpose it is easier to aim directly for $(0,1)$. Namely, if you have a different open interval $(a,b)$, you can map it bijectively to $(0,1)$ by a simple linear interpolation:
$$ x \in (a,b) \mapsto \frac{x-a}{b-a} \in (0,1) $$
This does not really specify a function. In order to have a function you need to have a particular $y$ for each $x\in(a,b)$, and a rule for which $y$s match up to which $x$s.