FRom https://proofwiki.org/wiki/Real_Numbers_are_Uncountable/Set-Theoretical_Approach:_Proof_1
- Why are we trying to construct an integer $e_o$ that's different from $d_{(0,0)}$ why exactly $e_o$ instead of $e_{62}$?
- Why specifically do we need an integer that's different and why exactly must it be different from $d_{(0,0)}$? instead of $d_{(42,12)}$?
- Why does the piecewise function from $e_m$ have a range of only 1 and 2? Why not have $e_m$ possibly be an integer from 0 through 9?
I feel like we're using a straw man's argument. We're forcing there to be no surjection. How do we know $e_0$ exists?
I feel like $e_m$ is a number in the diagonal of cantor's diagonalization argument.
Most importantly how do we even know $e_m$ exists?
It's a little unclear exactly what your problem is, and "I feel like $e_m$ is a number in the diagonal of cantor's diagonalization argument." doesn't make it better, this is Cantor's diagonalisation argument, so of course $e_m$ is a number from that.
We are trying to construct a real number that is different from any value of $f(n)$. We do that digit by digit, except that before the decimal point we consider all the (finitely many) digits at once, but it still makes sense to just select something that is different from the integer part of $f(0)$. If we selected $e_0$ to be different from $d_{(42,12)}$, we wouldn't be sure that the number we're generating isn't $f(0)$, which would ruin the proof.
We don't need more that two different values for each digit of the number we're constructing, in this proof the author has selected $1$ and $2$, but $4$ and $7$ would have worked just as well. Note that using $9$ opens for some speculation based on $0.999\ldots = 1$ (search the site if you want to know about that), so we can't completely say that any choice of digits work equally well.
We know that the $e_k$'s (for $k\in\mathbb N_0$) exists because we have given explicit formulas for each of them.