[gowld]

Then let U_i be the interval of length 1/3^i centered on f(i), so that the total length is 1/2, far less than 1.

Even though the supposed "surjection" is infinite, it's still the case that every x in [0,1] would be in in one of the finite U_n and therefore V_n. But every K_n clearly has measure > 0 and is therefore non-empty, and since the K_n are nested subsets, there is at least one special point x_omega that is in all of the K_n.

The "intuitive" problem (not logical problem) with PP's proof is that it relies on measure and completeness, which is far more technologically complex than the decimal diagonizalization argument.

Here is intuitive "rebuttal": the same proof strategy seemingly proves that the rationals are uncountable! (This is of course technically false, because rational intervals are incomplete and all have measure 0 in the first place. But understanding this is much more complicated than imagining an 2-D infinite spreadsheet of decimal numbers between 0 and 1.)

[wbl]

Are these the same proof?

If I let U_i be the interval of length 1/10^(i) centered on f(i), than what I'm saying is pick a different decimal digit to avoid this particular real.